Limits and Continuity of a Numerical Function — 2BAC PC / SVT
Interactive English course on limits and continuity: limits at a point and at infinity, indeterminate forms, composite limits, comparison and squeeze, continuity, extension by continuity, intermediate value theorem, bisection, inverse function, n-th roots and rational powers.
Mohamed Lagzouli
Créateur de ce parcours interactif.
Welcome
Welcome to this course
This course covers the first chapter of the 2BAC Physical Sciences and Life & Earth Sciences programme. It is designed to be followed in order: every result in Part C relies on Parts A and B.
How to work through this course
- Read one idea, then work its example before moving to the next.
- Look for each answer before revealing it — that is where the learning happens.
- Go back to the Core course as soon as an exercise resists.
Diagnostic: where do you stand?
Nine questions on the prerequisites. Answer first, reveal afterwards. If three or more answers escape you, review the limits from 1BAC before starting Part A.
1. Compute \(\displaystyle\lim_{x \to 2} (x^2 - 3x + 1)\).
Show the answer
A polynomial: substitute directly.
2. Let \(f(x) = \dfrac{x^2 - 1}{x - 1}\). Compute \(\displaystyle\lim_{x \to 1} f(x)\).
Show the answer
Indeterminate form \(\frac{0}{0}\). Factorise:
\[ \frac{x^2-1}{x-1} = \frac{(x-1)(x+1)}{x-1} = x+1 \quad (x \neq 1) \]So the limit is \(2\).
3. What is \(\displaystyle\lim_{x \to 0^+} \dfrac{1}{x}\)? And \(\displaystyle\lim_{x \to 0^-} \dfrac{1}{x}\)?
Show the answer
The two limits differ: the limit at \(0\) does not exist.
4. Give the domain of definition of \(g(x) = \sqrt{x - 3}\).
Show the answer
We need \(x - 3 \geqslant 0\), that is \(D_g = [3 ; +\infty[\).
5. Can a strictly increasing function on \([a;b]\) take the same value twice?
Show the answer
No. If \(x_1 < x_2\) then \(f(x_1) < f(x_2)\): two distinct points have distinct images. This remark will be the uniqueness argument throughout the chapter.
6. Translate: “the curve of \(f\) can be drawn without lifting the pen on \([0;1]\)”.
Show the answer
\(f\) is continuous on \([0;1]\). The picture is a correct one, but it is not a proof: at 2BAC level, continuity is established through the definition or through the theorems on operations.
Aim of the chapter. Move from “the curve looks continuous” to “I prove that \(f\) is continuous, then I draw a conclusion from it”.
Question 7. True or false: if \(\displaystyle\lim_{x \to 1^{-}} f(x) = -1\) and \(\displaystyle\lim_{x \to 1^{+}} f(x) = 1\), then \(f\) has a limit at \(1\).
Show the answer
False. Both limits exist but they differ.
A limit at a point exists only when the right-hand and left-hand limits are equal. This is the most common mistake of the chapter.
Question 8. What is the value of \(\infty - \infty\)?
Show the answer
It has no value at all. It is an indeterminate form.
\[ \lim_{x \to +\infty}\big((x+1)-x\big) = 1 \qquad \lim_{x \to +\infty}\big(x^2-x\big) = +\infty \]Two expressions of the same form, two different results: the writing must be transformed before concluding.
Question 9. The function \(x \mapsto \sin x\) is bounded. Does it have a limit at \(+\infty\)?
Show the answer
No. Being bounded is not enough to have a limit.
\(\sin(2k\pi) = 0\) and \(\sin\!\left(\frac{\pi}{2}+2k\pi\right) = 1\): the function does not get close to any single value.
Free preview
Here is the central idea of the chapter and its most useful application. The full course develops every point with its own examples.
Continuity at a point. Let \(f\) be defined on an interval \(I\) and let \(a \in I\). We say that \(f\) is continuous at \(a\) when
\[ \lim_{x \to a} f(x) = f(a) \]Three conditions are therefore met at once: \(f(a)\) exists, the limit exists, and the two coincide. If one of them fails, \(f\) is not continuous at \(a\).
Example. Let \(f(x) = \dfrac{x^2 - 4}{x - 2}\) for \(x \neq 2\) and \(f(2) = 4\). Let us study continuity at \(2\).
\[ \frac{x^2-4}{x-2} = \frac{(x-2)(x+2)}{x-2} = x+2 \quad (x \neq 2) \] \[ \lim_{x \to 2} f(x) = 4 = f(2) \]So \(f\) is continuous at \(2\).
Intermediate value theorem. If \(f\) is continuous on \([a;b]\), then for every real number \(k\) lying between \(f(a)\) and \(f(b)\), the equation \(f(x) = k\) has at least one solution in \([a;b]\).
Three questions to test yourself
1. Let \(f(x) = \dfrac{x^2 - 9}{x - 3}\) for \(x \neq 3\) and \(f(3) = 5\). Is \(f\) continuous at \(3\)?
Show the answer
No. \(\displaystyle\lim_{x \to 3} f(x) = 6\) whereas \(f(3) = 5\). The limit exists, the value exists, but they differ.
2. \(f\) is continuous on \([0;3]\), \(f(0) = -2\) and \(f(3) = 7\). What can be stated?
Show the answer
Since \(0\) lies between \(-2\) and \(7\), the equation \(f(x) = 0\) has at least one solution in \([0;3]\). Without a monotonicity assumption, nothing can be said about uniqueness.
3. Why can the IVT not be applied to \(f(x) = \dfrac{1}{x}\) on \([-1;1]\)?
Show the answer
\(f\) is not defined at \(0\), hence not continuous on \([-1;1]\). The hypothesis of the theorem fails, so the conclusion collapses. Indeed \(f(-1) = -1\) and \(f(1) = 1\), yet \(f\) never vanishes.
Remember. The IVT gives existence. Uniqueness requires an extra argument: strict monotonicity. That is the subject of the corollary, treated in Part C.
A limit that is computed after a transformation
What is \(\displaystyle\lim_{x \to 2} \frac{x^2 - 4}{x - 2}\)?
Direct substitution gives \(\frac{0}{0}\): an indeterminate form, which is not a result but a signal.
Show the answer
Factorise the numerator:
\[ \frac{x^2-4}{x-2} = \frac{(x-2)(x+2)}{x-2} = x+2 \quad (x \neq 2) \] \[ \lim_{x \to 2} \frac{x^2-4}{x-2} = 4 \]The function is not defined at \(2\), but the limit there is finite: it can be extended by continuity by setting \(f(2) = 4\).
The full course covers limits and then continuity, with nine worked examples, twelve exercises and two corrected exam papers.
Core course
Part A — Limits of a numerical function
L1. The four limit situations
Finite limit at a point
Let \(f\) be defined on an open interval containing \(a\), except possibly at \(a\). We say that \(f\) has \(\ell\) as its limit at \(a\) when the values \(f(x)\) become as close as we wish to \(\ell\) as soon as \(x\) is close enough to \(a\).
Right-hand limit and left-hand limit
The limit at \(a\) exists if and only if these two limits exist and are equal.
This is the most frequent mistake of the chapter. For \(f(x)=\dfrac{|x-1|}{x-1}\), we have \(\lim_{x\to 1^{-}}f(x)=-1\) and \(\lim_{x\to 1^{+}}f(x)=1\): both limits exist, they differ, so \(f\) has no limit at \(1\).
Infinite limit at a point
We say that \(f\) tends to \(+\infty\) as \(x\) tends to \(a\) when \(f(x)\) exceeds every number fixed in advance, however large, as soon as \(x\) is close enough to \(a\). We write:
Mind the wording: we say the limit is infinite, but the function then has no finite limit. Infinity is not a number.
Example 1 — an infinite limit, and the role of the sign
Study the behaviour of \(f(x)=\dfrac{1}{x}\) and then of \(g(x)=\dfrac{1}{x^{2}}\) at \(0\).
Show the solution
For \(g\), the denominator \(x^{2}\) is strictly positive on both sides of \(0\):
For \(f\), the denominator changes sign:
The two limits differ: \(f\) has no limit at \(0\), whereas \(g\) has one. Everything turns on the sign of the denominator.
Limit at infinity
Here we study the behaviour of \(f(x)\) as \(x\) becomes very large in absolute value.
A finite limit \(\ell\) at \(+\infty\) means that the curve gets indefinitely close to the horizontal line \(y=\ell\) without necessarily reaching it.
Standard limits
| Function (\(n\in\mathbb{N}^{*}\)) | At \(0^{+}\) | At \(+\infty\) | At \(-\infty\) |
|---|---|---|---|
| \(x^{n}\), \(n\) even | \(0\) | \(+\infty\) | \(+\infty\) |
| \(x^{n}\), \(n\) odd | \(0\) | \(+\infty\) | \(-\infty\) |
| \(\dfrac{1}{x^{n}}\) | \(+\infty\) | \(0\) | \(0\) |
| \(\sqrt{x}\) | \(0\) | \(+\infty\) | not defined |
Two trigonometric limits to know:
L2. Rules of computation
Sum of two limits
Here \(\ell\) and \(\ell'\) denote finite limits.
| \(\lim f\) | \(\lim g\) | \(\lim (f+g)\) |
|---|---|---|
| \(\ell\) | \(\ell'\) | \(\ell+\ell'\) |
| \(\ell\) | \(\pm\infty\) | \(\pm\infty\) |
| \(+\infty\) | \(+\infty\) | \(+\infty\) |
| \(-\infty\) | \(-\infty\) | \(-\infty\) |
| \(+\infty\) | \(-\infty\) | indeterminate |
Product of two limits
| \(\lim f\) | \(\lim g\) | \(\lim (f\times g)\) |
|---|---|---|
| \(\ell\) | \(\ell'\) | \(\ell\ell'\) |
| \(\ell>0\) | \(+\infty\) | \(+\infty\) |
| \(\ell<0\) | \(+\infty\) | \(-\infty\) |
| \(\pm\infty\) | \(\pm\infty\) | rule of signs |
| \(0\) | \(\pm\infty\) | indeterminate |
Quotient of two limits
| \(\lim f\) | \(\lim g\) | \(\lim \frac{f}{g}\) |
|---|---|---|
| \(\ell\) | \(\ell'\neq 0\) | \(\dfrac{\ell}{\ell'}\) |
| \(\ell\neq 0\) | \(0^{+}\) or \(0^{-}\) | \(\pm\infty\) according to the signs |
| \(\ell\) | \(\pm\infty\) | \(0\) |
| \(\pm\infty\) | \(\ell'\neq 0\) | \(\pm\infty\) according to the signs |
| \(0\) | \(0\) | indeterminate |
| \(\pm\infty\) | \(\pm\infty\) | indeterminate |
When the denominator tends to \(0\), the sign through which it vanishes decides the result. That is why in this case the right-hand and left-hand limits are always studied separately.
The four indeterminate forms:
None of these expressions equals \(0\), or \(1\), or \(\infty\). They merely signal that direct computation fails and that the expression must be transformed. Two indeterminate forms of the same type may give different results.
Resolving an indeterminate form
- Polynomial \(\frac{0}{0}\): factorise by \((x-a)\) in the numerator and in the denominator.
- Rational \(\frac{\infty}{\infty}\): factor out the highest-degree term.
- \(\infty-\infty\) with square roots: multiply by the conjugate expression.
- \(0\times\infty\): reduce it to a quotient.
Example 2 — form \(\frac{0}{0}\)
Show the solution
Numerator and denominator both vanish at \(2\). Factorise:
Example 3 — form \(\frac{\infty}{\infty}\)
Show the hint
Factor out \(x^{2}\) above and below.
Show the solution
Every term in \(\frac{1}{x}\) or \(\frac{1}{x^{2}}\) tends to \(0\), hence:
Example 4 — form \(\infty-\infty\)
Show the hint
Multiply and divide by the conjugate expression \(\sqrt{x^{2}+x}+x\).
Show the solution
For \(x>0\), factor \(x\) out of the denominator:
L3. Limit of a composite function
Limit of a composite
Let \(a\), \(b\) and \(c\) be finite or infinite, and suppose
These two hypotheses are not enough when \(b\) is finite. One of the following two conditions must be added.
First case — \(b\) is infinite. Nothing to add:
Second case — \(b\) is finite and \(v\) is continuous at \(b\). Then \(c=v(b)\) and the conclusion holds with no further hypothesis.
Third case — \(b\) is finite and \(v\) is not assumed continuous at \(b\). Then \(u(x)\neq b\) is required on a punctured neighbourhood of \(a\), that is, for every \(x\) close to \(a\) but different from \(a\).
In practice we set \(t=u(x)\): this is a change of variable. One must also check that \(u(x)\) belongs to the domain of \(v\) near \(a\), but that check alone is not enough to justify the passage to the limit.
Here is why the third condition is indispensable. Take:
Indeed \(\lim_{x\to 0}u(x)=0\) and \(\lim_{t\to 0}v(t)=1\). Yet \((v\circ u)(x)=v(0)=5\) for every \(x\), so the composite tends to \(5\), and not to \(1\).
The reason is that \(u\) takes exactly the value \(b=0\), precisely where \(v\) is discontinuous. In example 5 below the problem does not arise: \(v=\sqrt{\;}\) is continuous at \(4\), so we are in the second case.
Example 5 — change of variable
Show the solution
Set \(t=\dfrac{4x^{2}+1}{x^{2}+3}\). Factoring out \(x^{2}\):
The function \(t\mapsto\sqrt{t}\) is continuous at \(4\), with \(\sqrt{4}=2\): we are in the second case of the theorem, no extra hypothesis is required. Hence:
L4. Limits and order
Comparison theorem
Let \(f\) and \(g\) be such that \(f(x)\leqslant g(x)\) near \(a\).
- If \(\lim_{x\to a} f(x)=+\infty\), then \(\lim_{x\to a} g(x)=+\infty\).
- If \(\lim_{x\to a} g(x)=-\infty\), then \(\lim_{x\to a} f(x)=-\infty\).
A single lower bound suffices to push towards \(+\infty\); a single upper bound to push towards \(-\infty\).
Squeeze theorem
Let \(f\), \(g\), \(h\) be defined near \(a\), such that \(g(x)\leqslant f(x)\leqslant h(x)\) on that neighbourhood. If \(g\) and \(h\) have the same finite limit \(\ell\) at \(a\), then \(f\) has \(\ell\) as its limit at \(a\).
Example 6 — squeeze with the sine function
Show the hint
Start from \(-1\leqslant \sin x\leqslant 1\) and divide by \(x>0\).
Show the solution
For every \(x>0\):
Both bounds tend to \(0\) at \(+\infty\), hence:
A bounded function does not necessarily have a limit: \(x\mapsto \sin x\) is bounded and has no limit at \(+\infty\). The squeeze theorem concludes only if both bounds converge to the same value.
Part B — Continuity
A1. Continuity at a point
Let \(f\) be a function defined on an interval \(I\) and let \(a \in I\). \(f\) is continuous at \(a\) if and only if
\[ \lim_{x \to a} f(x) = f(a) \]Otherwise, \(f\) is discontinuous at \(a\).
The definition requires three things: that \(a\) belong to the domain, that the limit exist and be finite, and that the two values be equal. A discontinuity may therefore come from three distinct causes.
Example A1. Let \(f(x) = 3x - 1\). At \(a = 2\):
\[ \lim_{x \to 2} (3x-1) = 5 \quad \text{and} \quad f(2) = 5 \]So \(f\) is continuous at \(2\).
Example A1'. Let \(g(x) = \dfrac{1}{x-1}\) for \(x \neq 1\) and \(g(1) = 0\). At \(a = 1\):
\[ \lim_{x \to 1^+} g(x) = +\infty \]The limit is not finite: \(g\) is discontinuous at \(1\).
A2. Right-continuity, left-continuity
\(f\) is right-continuous at \(a\) if
\[ \lim_{x \to a^+} f(x) = f(a) \]\(f\) is left-continuous at \(a\) if
\[ \lim_{x \to a^-} f(x) = f(a) \]\(f\) is continuous at \(a\) if and only if it is right-continuous and left-continuous at \(a\):
\[ \lim_{x \to a^+} f(x) = \lim_{x \to a^-} f(x) = f(a) \]This criterion is the reference tool for piecewise-defined functions: each side of the matching point is studied separately.
Example A2. Let
\[ f(x) = \begin{cases} x + 1 & \text{if } x \leqslant 1 \\ 3 - x & \text{if } x > 1 \end{cases} \]Study at \(a = 1\):
\[ \lim_{x \to 1^-} f(x) = 1 + 1 = 2, \qquad \lim_{x \to 1^+} f(x) = 3 - 1 = 2, \qquad f(1) = 2 \]All three values coincide: \(f\) is continuous at \(1\).
A3. Continuity on an interval
\(f\) is continuous on an open interval \(]a;b[\) if it is continuous at every point of that interval.
\(f\) is continuous on \([a;b]\) if it is continuous on \(]a;b[\), right-continuous at \(a\) and left-continuous at \(b\).
At the ends of a closed interval, only continuity from the inner side makes sense — which is why the definition treats them separately.
A4. Continuity of the standard functions
The following functions are continuous on every interval contained in their domain:
- polynomial functions, on \(\mathbb{R}\);
- rational functions, on every interval where the denominator does not vanish;
- \(x \mapsto \sqrt{x}\), on \([0;+\infty[\);
- \(x \mapsto \sin x\) and \(x \mapsto \cos x\), on \(\mathbb{R}\);
- \(x \mapsto \tan x\), on every interval not containing \(\frac{\pi}{2} + k\pi\);
- \(x \mapsto |x|\), on \(\mathbb{R}\).
Practical consequence. We almost never go back to the definition. We identify the function as a standard one, or as built from standard ones by the operations of A5.
A5. Operations on continuous functions
Let \(f\) and \(g\) be continuous at \(a\), and let \(\lambda \in \mathbb{R}\). Then:
- \(f + g\) is continuous at \(a\);
- \(\lambda f\) is continuous at \(a\);
- \(f \times g\) is continuous at \(a\);
- \(\dfrac{f}{g}\) is continuous at \(a\) if \(g(a) \neq 0\);
- \(|f|\) is continuous at \(a\);
- \(\sqrt{f}\) is continuous at \(a\) if \(f(a) \geqslant 0\) and \(f \geqslant 0\) near \(a\).
Example A5. Let \(h(x) = \dfrac{x^2 + 1}{x - 3}\). On \(]3;+\infty[\): numerator and denominator are polynomials, hence continuous; the denominator does not vanish. So \(h\) is continuous on \(]3;+\infty[\). Same argument on \(]-\infty;3[\).
A6. Continuity of the composite
If \(u\) is continuous at \(a\) and \(v\) continuous at \(u(a)\), then \(v \circ u\) is continuous at \(a\).
If \(u\) is continuous on \(I\), \(u(I) \subset J\) and \(v\) continuous on \(J\), then \(v \circ u\) is continuous on \(I\).
Example A6. Let \(f(x) = \sqrt{x^2 + 1}\). Set \(u(x) = x^2 + 1\) and \(v(t) = \sqrt{t}\). \(u\) is continuous on \(\mathbb{R}\), and \(u(x) \geqslant 1 > 0\) so \(u(\mathbb{R}) \subset [1;+\infty[\) where \(v\) is continuous. Hence \(f = v \circ u\) is continuous on \(\mathbb{R}\).
A7. Image of an interval, image of a closed interval
If \(f\) is continuous on an interval \(I\), then \(f(I)\) is an interval.
If \(f\) is continuous on a closed interval \([a;b]\), then \(f([a;b])\) is a closed interval \([m;M]\), where \(m\) is the minimum and \(M\) the maximum of \(f\) on \([a;b]\). In particular, \(f\) is bounded and attains its bounds.
Careful. \(f([a;b]) = [m;M]\) — not \([f(a);f(b)]\). The equality \(f([a;b]) = [f(a);f(b)]\) holds only if \(f\) is continuous and increasing on \([a;b]\).
Example A7. Let \(f(x) = x^2\) on \([-1;2]\). \(f(-1) = 1\) and \(f(2) = 4\), but \(f([-1;2]) = [0;4]\): the minimum \(0\) is attained at \(x = 0\), inside the interval. The interval \([1;4]\) would be a wrong answer.
Monotonic case. If \(f\) is continuous and strictly monotonic on \(I\), the image is read at the ends:
- \(f\) increasing on \([a;b]\): \(f([a;b]) = [f(a);f(b)]\);
- \(f\) decreasing on \([a;b]\): \(f([a;b]) = [f(b);f(a)]\);
- \(f\) increasing on \([a;b[\): \(f([a;b[) = \left[f(a); \lim_{x \to b^-} f(x)\right[\).
A8. Extension by continuity
Let \(f\) be defined on \(I \setminus \{a\}\), where \(I\) is an interval containing \(a\). The function \(f\) is therefore not defined at \(a\), and the question of its continuity at that point does not arise as such.
We say that \(f\) can be extended by continuity at \(a\) when \(\displaystyle\lim_{x \to a} f(x)\) exists and is finite. Writing \(\ell\) for that limit, we then set:
\[ \tilde f(x) = \begin{cases} f(x) & \text{if } x \in I \setminus \{a\} \\ \ell & \text{if } x = a \end{cases} \]The function \(\tilde f\) so defined is continuous at \(a\): it is the extension of \(f\) by continuity.
Two conditions, and neither is optional.
- The limit must exist: if the right-hand and left-hand limits differ, no extension is possible.
- It must be finite: an infinite limit forbids any extension, since one cannot set \(\tilde f(a) = +\infty\).
When both hold, the value \(\ell\) is imposed: the extension exists and is unique. One does not choose \(\tilde f(a)\).
Example. Let \(f(x) = \dfrac{x^2 - 4}{x - 2}\), defined on \(\mathbb{R} \setminus \{2\}\). Can it be extended by continuity at \(2\)?
For \(x \neq 2\), factorise:
\[ \frac{x^2-4}{x-2} = \frac{(x-2)(x+2)}{x-2} = x+2 \] \[ \lim_{x \to 2} f(x) = 4 \]The limit exists and is finite: \(f\) can be extended by continuity at \(2\), by setting \(\tilde f(2) = 4\).
Counter-example. For \(g(x) = \dfrac{1}{x-2}\), we have \(\displaystyle\lim_{x \to 2^{+}} g(x) = +\infty\) and \(\displaystyle\lim_{x \to 2^{-}} g(x) = -\infty\). Neither existence nor finiteness holds: \(g\) cannot be extended by continuity at \(2\) in any way.
The notion comes back to A1: extending means precisely adding the missing value so that the three conditions for continuity at a point are finally met.
Part C — Fundamental theorems
B1. Intermediate value theorem
IVT. Let \(f\) be continuous on \([a;b]\). For every real \(k\) lying between \(f(a)\) and \(f(b)\), there exists at least one \(c \in [a;b]\) such that \(f(c) = k\).
Special case. If \(f\) is continuous on \([a;b]\) and \(f(a) \cdot f(b) < 0\), then the equation \(f(x) = 0\) has at least one solution in \(]a;b[\).
Example B1. Let us show that \(x^3 + x - 1 = 0\) has a solution in \([0;1]\). Set \(f(x) = x^3 + x - 1\), continuous on \(\mathbb{R}\) as a polynomial.
\[ f(0) = -1 < 0, \qquad f(1) = 1 > 0 \]\(f\) is continuous on \([0;1]\) and \(f(0) \cdot f(1) < 0\): by the IVT, there exists \(c \in ]0;1[\) such that \(f(c) = 0\).
Expected write-up. Three compulsory elements: continuity (with its justification), the interval, and the sign or the bracketing. Omitting the justification of continuity costs marks.
B2. Corollary: existence and uniqueness
If \(f\) is continuous and strictly monotonic on \([a;b]\), then for every \(k\) lying between \(f(a)\) and \(f(b)\), the equation \(f(x) = k\) has a unique solution in \([a;b]\).
The IVT provides existence, strict monotonicity provides uniqueness. These are two distinct arguments, to be cited separately.
Example B2. Take \(f(x) = x^3 + x - 1\) again on \([0;1]\). \(f\) is differentiable and \(f'(x) = 3x^2 + 1 > 0\), so \(f\) is strictly increasing on \([0;1]\). Together with example B1, the equation \(f(x) = 0\) has a unique solution \(\alpha \in ]0;1[\).
B3. Bracketing by bisection
Bisection brackets the solution \(\alpha\) by halving the interval at each step and keeping the half where the sign change persists.
Example B3. Let us bracket \(\alpha\), the solution of \(x^3 + x - 1 = 0\) in \([0;1]\), to within \(0{,}25\).
\[ f(0{,}5) = 0{,}125 + 0{,}5 - 1 = -0{,}375 < 0 \]So \(\alpha \in ]0{,}5 ; 1[\).
\[ f(0{,}75) = 0{,}421875 + 0{,}75 - 1 = 0{,}171875 > 0 \]So \(\alpha \in ]0{,}5 ; 0{,}75[\), of width \(0{,}25\).
B4. Inverse function
If \(f\) is continuous and strictly monotonic on an interval \(I\), then \(f\) is a bijection from \(I\) onto \(J = f(I)\). It therefore has an inverse function \(f^{-1} : J \to I\), which is:
- continuous on \(J\);
- strictly monotonic on \(J\), in the same direction as \(f\).
For every \(x \in I\) and every \(y \in J\):
\[ f^{-1}(f(x)) = x, \qquad f(f^{-1}(y)) = y \] \[ y = f(x) \iff x = f^{-1}(y) \]In an orthonormal frame, the curves of \(f\) and \(f^{-1}\) are symmetric about the line of equation \(y = x\).
Example B4. Let \(f(x) = x^2\) on \(I = [0;+\infty[\). \(f\) is continuous and strictly increasing on \(I\), and \(f(I) = [0;+\infty[\). So \(f\) is a bijection from \([0;+\infty[\) onto itself, and
\[ f^{-1}(y) = \sqrt{y} \quad \text{for } y \geqslant 0 \]On the whole of \(\mathbb{R}\), \(f\) is not injective (\(f(-2) = f(2)\)): restricting to \([0;+\infty[\) is indispensable.
B5. \(n\)-th root
Let \(n \in \mathbb{N}^*\). The function \(x \mapsto x^n\) is continuous and strictly increasing on \([0;+\infty[\), with image \([0;+\infty[\). Its inverse is the \(n\)-th root:
\[ \sqrt[n]{\;} : [0;+\infty[ \;\to\; [0;+\infty[ \]For \(x \geqslant 0\), \(\sqrt[n]{x}\) is the unique non-negative real whose \(n\)-th power equals \(x\):
\[ y = \sqrt[n]{x} \iff y \geqslant 0 \text{ and } y^n = x \]A note on the chosen setting. For odd \(n\), there is an \(n\)-th root defined on the whole of \(\mathbb{R}\): for instance \(-2\) satisfies \((-2)^3 = -8\), and one may legitimately write \(\sqrt[3]{-8} = -2\) in that wider setting.
In this course we consider the restriction to \([0;+\infty[\), whatever \(n\) may be. This is a presentation convention, not an impossibility: on that interval, \(x \mapsto x^n\) is bijective for every \(n\), and the properties above all hold simultaneously, which is no longer the case if one mixes even and odd exponents over \(\mathbb{R}\).
For \(a \geqslant 0\), \(b \geqslant 0\), \(n, m \in \mathbb{N}^*\):
\[ \sqrt[n]{ab} = \sqrt[n]{a} \cdot \sqrt[n]{b}, \qquad \sqrt[n]{\frac{a}{b}} = \frac{\sqrt[n]{a}}{\sqrt[n]{b}} \;(b > 0) \] \[ \sqrt[n]{a^m} = \left(\sqrt[n]{a}\right)^m, \qquad \sqrt[m]{\sqrt[n]{a}} = \sqrt[mn]{a} \]The function \(x \mapsto \sqrt[n]{x}\) is continuous on \([0;+\infty[\).
Example B5. Let us simplify \(\sqrt[3]{8} \cdot \sqrt[3]{27}\):
\[ \sqrt[3]{8} \cdot \sqrt[3]{27} = \sqrt[3]{8 \times 27} = \sqrt[3]{216} = 6 \]Check: \(2 \times 3 = 6\).
B6. Rational powers
For \(a > 0\), \(p \in \mathbb{Z}\) and \(q \in \mathbb{N}^*\), we set
\[ a^{\frac{p}{q}} = \sqrt[q]{a^p} \]In particular \(a^{\frac{1}{n}} = \sqrt[n]{a}\).
For \(a > 0\), \(b > 0\) and \(r, s \in \mathbb{Q}\):
\[ a^r \cdot a^s = a^{r+s}, \qquad \frac{a^r}{a^s} = a^{r-s}, \qquad (a^r)^s = a^{rs} \] \[ (ab)^r = a^r b^r, \qquad a^{-r} = \frac{1}{a^r} \]Example B6. Let us compute \(8^{\frac{2}{3}}\):
\[ 8^{\frac{2}{3}} = \sqrt[3]{8^2} = \sqrt[3]{64} = 4 \]Or \(8^{\frac{2}{3}} = \left(\sqrt[3]{8}\right)^2 = 2^2 = 4\) — both routes give the same result.
Chapter summary
- Proving continuity: standard functions + operations + composition. The definition only comes in at matching points.
- Piecewise function: right-hand limit = left-hand limit = value.
- Image of a closed interval: \([m;M]\), not \([f(a);f(b)]\), unless monotonic.
- Existence of a solution: IVT, with continuity justified.
- Uniqueness: strict monotonicity, a separate argument.
- Inverse: continuous, strictly monotonic in the same direction, curves symmetric about \(y = x\).
Worked examples
Nine situations written out step by step. Look for each step before revealing it.
Example 1 — Continuity of a piecewise function
Let
\[ f(x) = \begin{cases} \dfrac{x^2 - 4}{x - 2} & \text{if } x \neq 2 \\ m & \text{if } x = 2 \end{cases} \]Find \(m\) so that \(f\) is continuous at \(2\).
Show the answer
Step 1 — Compute the limit. For \(x \neq 2\):
\[ \frac{x^2-4}{x-2} = \frac{(x-2)(x+2)}{x-2} = x+2 \] \[ \lim_{x \to 2} f(x) = 4 \]Step 2 — Apply the definition. \(f\) continuous at \(2\) is equivalent to \(\lim_{x \to 2} f(x) = f(2)\), hence \(4 = m\).
Conclusion. \(m = 4\).
Example 2 — Matching from the right and from the left
Let
\[ g(x) = \begin{cases} x^2 + 1 & \text{if } x < 1 \\ ax + b & \text{if } x \geqslant 1 \end{cases} \]We know that \(g\) is continuous at \(1\) and that \(g(3) = 8\). Find \(a\) and \(b\).
Show the answer
Step 1 — Continuity at 1.
\[ \lim_{x \to 1^-} g(x) = 1 + 1 = 2, \qquad \lim_{x \to 1^+} g(x) = a + b = g(1) \]Continuity forces \(a + b = 2\).
Step 2 — Use \(g(3) = 8\). Since \(3 \geqslant 1\):
\[ 3a + b = 8 \]Step 3 — Solve. Subtracting: \(2a = 6\), hence \(a = 3\) and \(b = -1\).
Check. \(a + b = 2\) ✓ and \(3(3) - 1 = 8\) ✓
Example 3 — Image of an interval
Let \(f(x) = x^2 - 4x + 1\) on \([0;3]\). Determine \(f([0;3])\).
Show the answer
Step 1 — Continuity. \(f\) is a polynomial, hence continuous on \([0;3]\). Its image is a closed interval \([m;M]\).
Step 2 — Variations. \(f'(x) = 2x - 4\), zero at \(x = 2\), negative on \([0;2[\), positive on \(]2;3]\). So \(f\) decreases then increases: the minimum is reached at \(2\).
\[ f(2) = 4 - 8 + 1 = -3 \]Step 3 — Compare the endpoints.
\[ f(0) = 1, \qquad f(3) = 9 - 12 + 1 = -2 \]The maximum is \(\max(1; -2) = 1\).
Conclusion. \(f([0;3]) = [-3 ; 1]\).
Trap avoided. Answering \([f(0);f(3)] = [-2;1]\) would be wrong: \(f\) is not monotonic on \([0;3]\), and the minimum \(-3\) is interior.
Example 4 — Existence and uniqueness of a solution
Let \(f(x) = x^3 + 3x - 5\). Show that the equation \(f(x) = 0\) has a unique solution \(\alpha\) in \(\mathbb{R}\), then locate it between two consecutive integers.
Show the answer
Step 1 — Continuity. \(f\) is a polynomial, hence continuous on \(\mathbb{R}\).
Step 2 — Strict monotonicity.
\[ f'(x) = 3x^2 + 3 = 3(x^2 + 1) > 0 \quad \text{for all } x \]So \(f\) is strictly increasing on \(\mathbb{R}\).
Step 3 — Limits at the ends.
\[ \lim_{x \to -\infty} f(x) = -\infty, \qquad \lim_{x \to +\infty} f(x) = +\infty \]Hence \(f(\mathbb{R}) = \mathbb{R}\), which contains \(0\).
Step 4 — Conclude. \(f\) is continuous and strictly increasing from \(\mathbb{R}\) onto \(\mathbb{R}\): it is a bijection. The equation \(f(x) = 0\) therefore has a unique solution \(\alpha\).
Step 5 — Locate it.
\[ f(1) = 1 + 3 - 5 = -1 < 0, \qquad f(2) = 8 + 6 - 5 = 9 > 0 \]Hence \(1 < \alpha < 2\).
Example 5 — Explicit inverse function
Let \(f(x) = \dfrac{2x + 1}{x - 1}\) defined on \(I = ]1;+\infty[\). Show that \(f\) is a bijection from \(I\) onto an interval \(J\) to be determined, then give \(f^{-1}\) explicitly.
Show the answer
Step 1 — Continuity. \(f\) is a rational function whose denominator does not vanish on \(]1;+\infty[\): \(f\) is continuous on \(I\).
Step 2 — Monotonicity.
\[ f'(x) = \frac{2(x-1) - (2x+1)}{(x-1)^2} = \frac{-3}{(x-1)^2} < 0 \]So \(f\) is strictly decreasing on \(I\).
Step 3 — Image.
\[ \lim_{x \to 1^+} f(x) = +\infty, \qquad \lim_{x \to +\infty} f(x) = 2 \]\(f\) decreasing and continuous on \(]1;+\infty[\) gives \(J = f(I) = ]2;+\infty[\).
Step 4 — Bijection. Continuous and strictly monotonic: \(f\) is a bijection from \(]1;+\infty[\) onto \(]2;+\infty[\).
Step 5 — Make it explicit. Solve \(y = \dfrac{2x+1}{x-1}\):
\[ y(x-1) = 2x+1 \;\Longrightarrow\; yx - y = 2x + 1 \;\Longrightarrow\; x(y-2) = y+1 \] \[ x = \frac{y+1}{y-2} \]Hence \(f^{-1}(y) = \dfrac{y+1}{y-2}\) for \(y \in ]2;+\infty[\).
Check. \(f(3) = \frac{7}{2}\) and \(f^{-1}\!\left(\frac{7}{2}\right) = \frac{9/2}{3/2} = 3\) ✓
Example 6 — Form \(\frac{0}{0}\): factorise
Compute \(\displaystyle\lim_{x \to 3} \frac{x^2 - 9}{x^2 - 4x + 3}\).
Show the hint
\(3\) cancels both polynomials: it is a root of each.
Show the solution
Example 7 — Form \(\frac{\infty}{\infty}\): highest degree
Compute \(\displaystyle\lim_{x \to -\infty} \frac{2x^3 - x}{5x^3 + 4x^2}\).
Show the solution
The result is the same at \(-\infty\) and at \(+\infty\): for a rational function whose degrees are equal, the limit is the ratio of the leading coefficients, on either side.
Example 8 — Form \(\infty - \infty\): conjugate expression
Compute \(\displaystyle\lim_{x \to +\infty}\left(\sqrt{x^2+3x} - x\right)\).
Show the hint
Multiply and divide by \(\sqrt{x^2+3x}+x\).
Show the solution
For \(x > 0\), we factored \(x\) out of the denominator.
Example 9 — Composition: the condition that gets forgotten
We know that \(\displaystyle\lim_{x \to a} u(x) = b\) and \(\displaystyle\lim_{t \to b} v(t) = c\). Can we conclude that \(\displaystyle\lim_{x \to a}(v \circ u)(x) = c\)?
Show the hint
Try with \(u\) constant equal to \(b\), and \(v\) discontinuous at \(b\).
Show the solution
Not in general. Take \(a=b=0\), \(u(x)=0\) for all \(x\), and
\[ v(t) = \begin{cases} 1 & \text{if } t \neq 0 \\ 5 & \text{if } t = 0 \end{cases} \]Both hypotheses hold, with \(c = 1\). Yet \((v \circ u)(x) = v(0) = 5\) for every \(x\).
When \(b\) is finite, one of these two conditions must be added:
- \(v\) is continuous at \(b\) — then \(c = v(b)\);
- \(u(x) \neq b\) on a punctured neighbourhood of \(a\).
When \(b\) is infinite, no extra condition is required.
Methods used
- Example 1 — Extension by continuity: compute the limit, then impose it as the value.
- Example 2 — Matching: two one-sided limits, then use of an additional condition. A system with two unknowns.
- Example 3 — Image of a closed interval: justify continuity, draw the table of variations, compare all the candidate values. Never settle for the endpoints alone.
- Example 4 — Existence + uniqueness: IVT then strict monotonicity, two distinct arguments. The bracketing is obtained by computing signs.
- Example 5 — Inverse: continuity, monotonicity, image, then algebraic solving of \(y = f(x)\) for \(x\).
Three mistakes not to repeat
- Writing \(f([a;b]) = [f(a);f(b)]\) without having established monotonicity.
- Concluding uniqueness from the IVT alone.
- Forgetting to justify continuity before invoking the IVT — it is the hypothesis of the theorem.
Methods used — examples 6 to 9
Example 6 — Form \(\frac{0}{0}\): factorise by \((x-a)\) in the numerator and in the denominator, then simplify for \(x \neq a\).
Example 7 — Form \(\frac{\infty}{\infty}\): factor out the highest-degree term above and below.
Example 8 — Form \(\infty - \infty\) with a square root: multiply by the conjugate expression, then factor out \(x\).
Example 9 — Composition: when \(b\) is finite, check that \(v\) is continuous at \(b\) or that \(u\) does not take the value \(b\) near \(a\).
Two mistakes not to repeat — limits part
Writing \(\infty-\infty = 0\) or \(\frac{\infty}{\infty} = 1\): these expressions denote no number at all.
Justifying a change of variable solely because \(u(x)\) belongs to the domain of \(v\): the condition is about continuity, or about the punctured neighbourhood.
Interactive exercises
Four levels. Work each question on paper before revealing the answer. The hints are there to unblock you, not to replace the search.
Level 1 — Direct application
Exercise 1. Study the continuity at \(0\) of
\[ f(x) = \begin{cases} \dfrac{\sqrt{x+1} - 1}{x} & \text{if } x \neq 0, \; x > -1 \\ \dfrac{1}{2} & \text{if } x = 0 \end{cases} \]a) Simplify the expression of \(f(x)\) for \(x \neq 0\).
Show the hint
Multiply numerator and denominator by \(\sqrt{x+1}+1\).
Show the answer
Multiply by the conjugate quantity:
\[ \frac{\sqrt{x+1}-1}{x} = \frac{(x+1)-1}{x\left(\sqrt{x+1}+1\right)} = \frac{1}{\sqrt{x+1}+1} \]Cancelling \(x\) is valid because \(x \neq 0\).
b) Deduce \(\displaystyle\lim_{x \to 0} f(x)\).
Show the answer
The initial \(\frac{0}{0}\) form is resolved: the simplified expression is continuous at \(0\).
c) Conclude about the continuity of \(f\) at \(0\).
Show the answer
All three conditions hold: \(f(0)\) exists, the limit exists, and the two coincide. So \(f\) is continuous at \(0\).
Exercise 2. Let \(h(x) = \dfrac{\sqrt{x}}{x^2 + 1}\).
a) Determine the domain of definition of \(h\).
Show the answer
The square root forces \(x \geqslant 0\). The denominator \(x^2+1\) never vanishes since \(x^2+1 \geqslant 1\). So \(D_h = [0;+\infty[\).
b) Justify the continuity of the numerator and that of the denominator.
Show the answer
\(x \mapsto \sqrt{x}\) is continuous on \([0;+\infty[\). \(x \mapsto x^2+1\) is a polynomial, hence continuous on \(\mathbb{R}\).
c) Conclude that \(h\) is continuous on \([0;+\infty[\).
Show the answer
The quotient of a continuous function by a continuous function that does not vanish is continuous. Since \(x^2+1 \geqslant 1 > 0\) on \([0;+\infty[\), \(h\) is continuous on \([0;+\infty[\).
Point of method. The condition to check is not “the denominator is continuous”, but “the denominator does not vanish”.
Exercise 3. Computing roots and rational powers.
a) Compute \(\sqrt[4]{16}\).
Show the answer
b) Compute \(27^{\frac{1}{3}}\).
Show the answer
c) Compute \(32^{\frac{2}{5}}\).
Show the answer
d) Which property allows the writing \(\left(\sqrt[5]{32}\right)^2 = \sqrt[5]{32^2}\), and under which conditions?
Show the answer
Valid for \(x \geqslant 0\) when \(m \in \mathbb{N}^{*}\), and for \(x > 0\) when \(m \leqslant 0\). The pair \(x = 0\), \(m = 0\) is excluded: it would lead to \(0^{0}\).
Here \(x = 32 > 0\) and \(m = 2\): the two writings are equivalent.
Exercise 9. Standard limits and operations.
a) Compute \(\displaystyle\lim_{x \to +\infty} \frac{3x^2 - x + 1}{2x^2 + 5}\).
Show the hint
Factor out \(x^2\) in the numerator and in the denominator.
Show the answer
b) Compute \(\displaystyle\lim_{x \to 0^{+}} \frac{1}{x}\) then \(\displaystyle\lim_{x \to 0^{-}} \frac{1}{x}\). What can be said about \(\displaystyle\lim_{x \to 0} \frac{1}{x}\)?
Show the answer
Both limits exist but they differ: \(\frac{1}{x}\) has no limit at \(0\).
c) Same question for \(\dfrac{1}{x^2}\). Explain the difference.
Show the answer
The denominator \(x^2\) is strictly positive on both sides, whereas \(x\) changes sign. Everything turns on the sign of the denominator.
d) Compute \(\displaystyle\lim_{x \to 0} \frac{\sin(3x)}{x}\).
Show the answer
We reduce to \(\displaystyle\lim_{t \to 0}\frac{\sin t}{t} = 1\) by setting \(t = 3x\).
Level 2 — Reasoning
Exercise 4. Let \(f(x) = x^3 - 3x + 1\).
a) Show that \(f(x) = 0\) has a solution in \([0;1]\).
Show the answer
\(f\) is a polynomial, hence continuous on \([0;1]\).
\[ f(0) = 1 > 0, \qquad f(1) = 1 - 3 + 1 = -1 < 0 \]\(f(0) \cdot f(1) < 0\): by the IVT, \(f\) vanishes at least once in \(]0;1[\).
b) Is this solution unique in \(\mathbb{R}\)?
Show the answer
No. \(f'(x) = 3x^2 - 3 = 3(x-1)(x+1)\) vanishes at \(-1\) and \(1\): \(f\) is not monotonic on \(\mathbb{R}\).
\[ f(-2) = -8 + 6 + 1 = -1 < 0, \qquad f(0) = 1 > 0, \qquad f(2) = 8 - 6 + 1 = 3 > 0 \]Three sign changes: the equation has three real solutions.
c) Is the solution in \([0;1]\) unique within that interval?
Show the hint
What is the sign of \(f'\) on \([0;1]\)?
Show the answer
On \([0;1]\), \(f'(x) = 3(x-1)(x+1) \leqslant 0\), and vanishes only at \(1\): \(f\) is strictly decreasing on \([0;1]\).
Existence comes from the IVT, and strict monotonicity makes it possible to guarantee uniqueness on that interval. That is the whole gap with question b): uniqueness depends on the interval one works in.
d) Bracket this solution by bisection, with a width of \(0{,}25\).
Show the answer
Initial interval \([a_0;b_0] = [0;1]\), of width \(1\).
Step 1 — \(m = 0{,}5\), \(f(0{,}5) = 0{,}125 - 1{,}5 + 1 = -0{,}375 \neq 0\), of opposite sign to \(f(0)\): the current interval becomes \([0;0{,}5]\).
Step 2 — \(m = 0{,}25\), \(f(0{,}25) \approx 0{,}266 \neq 0\), of the same sign as \(f(0)\): the current interval becomes \([0{,}25;0{,}5]\).
After \(k = 2\) steps, the width is indeed:
\[ \frac{b_0 - a_0}{2^{k}} = \frac{1}{2^{2}} = 0{,}25 \]Had \(f(m)\) been exactly zero at some step, the algorithm would have stopped: \(m\) would have been the exact solution.
Exercise 5. Let \(f(x) = x^2 + 2x\) on \([-2;1]\).
a) Justify that \(f([-2;1])\) is a closed interval.
Show the answer
\(f\) is a polynomial, hence continuous. The image of a closed interval under a continuous function is a closed interval.
b) Study the variations of \(f\) on \([-2;1]\).
Show the answer
\(f'(x) = 2x + 2\), zero at \(x = -1\), negative before, positive after.
\(f\) decreases on \([-2;-1]\) then increases on \([-1;1]\): it is not monotonic.
c) Deduce \(f([-2;1])\).
Show the answer
Maximum \(= 3\). So \(f([-2;1]) = [-1;3]\).
Mistake to avoid. Reading the image from the endpoints alone would give \([0;3]\): that is wrong, because \(f\) is not monotonic on this interval.
Exercise 6. Find \(a\) and \(b\) so that
\[ f(x) = \begin{cases} ax + 2 & \text{if } x \leqslant 0 \\ \dfrac{\sin(bx)}{x} & \text{if } x > 0 \end{cases} \]is continuous at \(0\), with \(b \neq 0\).
a) Compute \(\displaystyle\lim_{x \to 0^{-}} f(x)\) and \(f(0)\).
Show the answer
For \(x \leqslant 0\), \(f(x) = ax + 2\), so \(f(0) = 2\) and
\[ \lim_{x \to 0^{-}} f(x) = 2 \]b) Compute \(\displaystyle\lim_{x \to 0^{+}} f(x)\) in terms of \(b\).
Show the hint
Bring out \(\dfrac{\sin(bx)}{bx}\).
Show the answer
c) Conclude about \(a\) and about \(b\).
Show the answer
Continuity forces the two limits and \(f(0)\) to be equal, hence \(b = 2\).
The value of \(a\) is arbitrary: it plays no part in the left-hand limit, which equals \(a \times 0 + 2 = 2\) independently of \(a\).
Point of method. A matching condition does not always determine every parameter. One must check which ones actually play a part.
Exercise 10. Indeterminate forms.
a) Compute \(\displaystyle\lim_{x \to 2} \frac{x^2 - 4}{x - 2}\).
Show the answer
Form \(\frac{0}{0}\) resolved by factorisation.
b) Compute \(\displaystyle\lim_{x \to +\infty}\left(\sqrt{x^2+x} - x\right)\).
Show the hint
Multiply by the conjugate expression \(\sqrt{x^2+x} + x\).
Show the answer
Form \(\infty - \infty\) resolved by the conjugate quantity.
c) Compute \(\displaystyle\lim_{x \to 0} \frac{x^2}{\sin x}\).
Show the answer
Form \(\frac{0}{0}\) reduced to a product one factor of which tends to \(0\).
d) Can two forms \(\infty - \infty\) have different limits? Give two examples.
Show the answer
Yes, and that is the whole meaning of the word “indeterminate”.
\[ \lim_{x \to +\infty}\big((x+1) - x\big) = 1 \qquad \lim_{x \to +\infty}\big((x^2) - x\big) = +\infty \]An indeterminate form is not a result: it is the signal that the expression must be transformed.
Level 3 — Comparison and squeeze
Exercise 11. Bounded functions, comparison and squeeze.
a) Bound \(\sin x\) for every real \(x\), then bound \(\dfrac{\sin x}{x}\) for \(x > 0\).
Show the answer
Dividing by \(x > 0\), which does not reverse the inequalities:
\[ -\frac{1}{x} \leqslant \frac{\sin x}{x} \leqslant \frac{1}{x} \]b) Deduce \(\displaystyle\lim_{x \to +\infty} \frac{\sin x}{x}\).
Show the answer
Both bounds \(-\frac{1}{x}\) and \(\frac{1}{x}\) tend to \(0\) at \(+\infty\).
\[ \lim_{x \to +\infty} \frac{\sin x}{x} = 0 \]This is the squeeze theorem: it concludes because both bounds converge to the same finite limit.
c) The function \(x \mapsto \sin x\) is bounded. Does it have a limit at \(+\infty\)? Justify.
Show the hint
Look at the values taken at \(x = 2k\pi\) and at \(x = \frac{\pi}{2} + 2k\pi\), for integer \(k\) tending to infinity.
Show the answer
No. Set \(x_k = 2k\pi\) and \(y_k = \frac{\pi}{2} + 2k\pi\).
\[ \sin(x_k) = 0 \quad \text{and} \quad \sin(y_k) = 1 \quad \text{for every } k \]Both sequences tend to \(+\infty\) but the values of \(\sin\) there stay equal to \(0\) and to \(1\) respectively. The function can therefore get close to no single value.
Remember. Being bounded is not enough to have a limit. What makes \(\frac{\sin x}{x}\) converge is not that \(\sin\) is bounded, it is that \(\frac{1}{x}\) tends to \(0\) and crushes the oscillation.
d) Let \(f\) be such that \(f(x) \geqslant x - \cos x\) for every \(x > 0\). Determine \(\displaystyle\lim_{x \to +\infty} f(x)\).
Show the answer
Since \(\cos x \leqslant 1\), we have \(x - \cos x \geqslant x - 1\), hence
\[ f(x) \geqslant x - 1 \quad \text{for every } x > 0 \]And \(x - 1 \to +\infty\). By the comparison theorem:
\[ \lim_{x \to +\infty} f(x) = +\infty \]Difference with b). Here a single lower bound is enough, because we are pushing towards \(+\infty\). The squeeze theorem requires two bounds and a finite limit.
Level 4 — Going further
Exercise 7. Let \(f\) be continuous on \([0;1]\) with \(f(0) = 0\) and \(f(1) = 1\).
a) We look for \(c\) such that \(f(c) = 1 - c\). Which auxiliary function should be introduced?
Show the answer
Set \(g(x) = f(x) - 1 + x\). The equation \(f(c) = 1-c\) is equivalent to \(g(c) = 0\), which turns the problem into a vanishing problem.
\(g\) is continuous on \([0;1]\) as a sum of continuous functions.
b) Compute \(g(0)\) and \(g(1)\).
Show the answer
c) Conclude, and extract the general method.
Show the answer
By the IVT, there exists \(c \in ]0;1[\) such that \(g(c) = 0\), that is \(f(c) = 1 - c\).
Method to remember. For an equation of the form \(f(x) = \varphi(x)\), introduce \(g = f - \varphi\) and apply the IVT to \(g\).
Note that nothing can be said about uniqueness: \(f\) is not assumed monotonic.
Exercise 8. Let \(f(x) = \dfrac{x}{x^2+1}\) on \([0;1]\).
a) Show that \(f\) is continuous and strictly increasing on \([0;1]\).
Show the answer
\(f\) is rational and \(x^2+1\) does not vanish: \(f\) is continuous on \([0;1]\).
\[ f'(x) = \frac{(x^2+1) - x(2x)}{(x^2+1)^2} = \frac{1-x^2}{(x^2+1)^2} \]On \([0;1[\), \(1 - x^2 > 0\) so \(f' > 0\): \(f\) is strictly increasing on \([0;1]\).
b) Deduce that \(f\) is a bijection from \([0;1]\) onto an interval \(J\) to be determined.
Show the answer
Continuous and strictly increasing on \([0;1]\), \(f\) is a bijection from \([0;1]\) onto \(J = \left[0 ; \tfrac{1}{2}\right]\).
c) Determine \(f^{-1}(y)\) for \(y \in J\).
Show the hint
Solve \(y = \dfrac{x}{x^2+1}\) for \(x\): it is a quadratic equation.
Show the answer
Let us solve \(y = \dfrac{x}{x^2+1}\):
\[ yx^2 - x + y = 0 \]For \(y \neq 0\), the discriminant is \(\Delta = 1 - 4y^2 \geqslant 0\) since \(y \leqslant \tfrac{1}{2}\). The roots are
\[ x = \frac{1 \pm \sqrt{1-4y^2}}{2y} \]For \(0 < y < \tfrac{1}{2}\), we have \(\Delta = 1-4y^2 > 0\): the two roots are distinct. Their product equals \(\dfrac{y}{y} = 1\), so one lies in \(]0;1[\) and the other in \(]1;+\infty[\). Only the branch with the \(-\) sign belongs to \([0;1]\):
\[ f^{-1}(y) = \frac{1 - \sqrt{1-4y^2}}{2y} \qquad \left(0 < y < \tfrac{1}{2}\right) \]At the two endpoints, the reasoning above does not apply and each case must be treated separately.
- \(y = 0\): the equation \(yx^2 - x + y = 0\) becomes \(-x = 0\). It is no longer quadratic, and the formula above is undefined. The only solution is \(x = 0\), so \(f^{-1}(0) = 0\).
- \(y = \tfrac{1}{2}\): then \(\Delta = 1 - 4 \times \tfrac14 = 0\). Both signs give the same value \(x = \dfrac{1}{2y} = 1\): there is no choice left, the root is double. So \(f^{-1}\!\left(\tfrac12\right) = 1\), which is consistent with \(f(1) = \tfrac12\).
Point of rigour. Writing “only the \(-\) branch works” is exact only as long as \(\Delta > 0\). At \(y = \tfrac12\) the two branches coincide, and at \(y = 0\) the equation changes nature: the endpoints of an interval always deserve their own examination.
Exercise 12. Limit of a composite function.
a) Compute \(\displaystyle\lim_{x \to +\infty} \sqrt{\frac{4x^2+1}{x^2+3}}\) and justify the passage to the limit.
Show the answer
Set \(t = \dfrac{4x^2+1}{x^2+3}\). Factoring out \(x^2\):
\[ \lim_{x \to +\infty} t = \frac{4}{1} = 4 \]The function \(t \mapsto \sqrt{t}\) is continuous at \(4\), with \(\sqrt{4}=2\): the passage to the limit is valid with no extra hypothesis.
\[ \lim_{x \to +\infty} \sqrt{\frac{4x^2+1}{x^2+3}} = 2 \]b) Let \(u(x) = 0\) for every \(x\), and let \(v\) be defined by \(v(t) = 1\) if \(t \neq 0\), \(v(0) = 5\). Compute \(\displaystyle\lim_{x \to 0} u(x)\), \(\displaystyle\lim_{t \to 0} v(t)\), then \(\displaystyle\lim_{x \to 0}(v \circ u)(x)\).
Show the answer
But \((v \circ u)(x) = v(0) = 5\) for every \(x\), hence
\[ \lim_{x \to 0} (v \circ u)(x) = 5 \neq 1 \]c) Which hypothesis is missing in question b), and why does question a) not need it?
Show the answer
When \(b\) is finite, the two limits are not enough. One must add, at choice: that \(v\) be continuous at \(b\), or that \(u(x) \neq b\) on a punctured neighbourhood of \(a\).
In b), \(u\) takes exactly the value \(b = 0\), precisely where \(v\) is discontinuous: neither condition holds.
In a), \(v = \sqrt{\;}\) is continuous at \(b = 4\): the first condition is satisfied, nothing else needs checking.
Point of method. Checking that \(u(x)\) belongs to the domain of \(v\) is not enough to justify a change of variable inside a limit.
Indicative mark scheme and points of method
| Exercise | Key method | Marks |
|---|---|---|
| 1 | Conjugate quantity, then the definition of continuity | 2 |
| 2 | Quotient of continuous functions, non-vanishing denominator | 2 |
| 3 | \(n\)-th roots and rational powers | 2 |
| 4 | IVT; counter-example to uniqueness through the sign of \(f'\) | 3 |
| 5 | Image of a closed interval through the table of variations | 3 |
| 6 | One-sided limits; \(\lim \frac{\sin t}{t} = 1\) | 3 |
| 7 | Auxiliary function \(g = f - \varphi\), then IVT | 2 |
| 8 | Bijection, image, algebraic inversion, selecting the root | 3 |
Targeted remediation
- Exercises 1, 2 or 3 missed → go back to the Core course, sections A1 to A5, then worked examples 1 and 2.
- Exercise 4 or 5 missed → review A7 and B1, then worked example 3. The mistake is almost always \(f([a;b]) = [f(a);f(b)]\) applied without monotonicity.
- Exercise 6 missed → review A2 and the trigonometric limits from 1BAC.
- Exercises 7 or 8 missed → review B2 and B4, then worked example 5.
Short answers
- 1. Continuous at \(0\).
- 2. Continuous on \([0;+\infty[\).
- 3. \(2\); \(3\); \(4\).
- 4. a) IVT on \([0;1]\). b) No, three solutions.
- 5. \([-1;3]\).
- 6. \(b = 2\), \(a\) arbitrary.
- 7. Set \(g(x) = f(x) - 1 + x\).
- 8. \(J = \left[0;\tfrac12\right]\); \(f^{-1}(y) = \dfrac{1-\sqrt{1-4y^2}}{2y}\).
Indicative mark scheme and points of method — exercises 9 to 12
Exercise 9 — Standard limits, sign of the denominator, trigonometric limit. 4 marks.
Exercise 10 — The four indeterminate forms and the techniques that resolve them. 4 marks.
Exercise 11 — Squeeze, comparison, and the counter-example of the bounded function with no limit. 4 marks.
Exercise 12 — Composition: continuity condition or punctured-neighbourhood condition. 3 marks.
Targeted remediation — limits part
If exercise 9 b) or c) resists: review the sign of the denominator when it tends to \(0\), and go back to example 1 of the course.
If exercise 10 resists: go back to the method “Resolving an indeterminate form” in the course, technique by technique, before redoing examples 6 to 8.
If exercise 11 c) resists: remember that being bounded gives no limit at all, and that the squeeze theorem requires two bounds converging to the same finite value.
If exercise 12 c) resists: redo example 9, which bears on exactly this point.
Short answers — exercises 9 to 12
9 a) \(\frac{3}{2}\) — b) \(+\infty\) and \(-\infty\), no limit — c) \(+\infty\) — d) \(3\).
10 a) \(4\) — b) \(\frac{1}{2}\) — c) \(0\) — d) \(1\) and \(+\infty\).
11 a) \(-\frac{1}{x} \leqslant \frac{\sin x}{x} \leqslant \frac{1}{x}\) — b) \(0\) — c) no — d) \(+\infty\).
12 a) \(2\) — b) \(0\), \(1\), then \(5\) — c) continuity at \(b\), or punctured neighbourhood.
Exam mode
Paper 1 — Continuity
\(1\)-hour paper, \(20\) marks. No documents allowed. Answer the three exercises in any order, writing out your reasoning in full.
Exercise I — Continuity and matching (6 marks)
Let
\[ f(x) = \begin{cases} \dfrac{x^2 - x - 2}{x - 2} & \text{if } x > 2 \\ ax + 1 & \text{if } x \leqslant 2 \end{cases} \]- 1. Compute \(\displaystyle\lim_{x \to 2^+} f(x)\). (2 marks)
- 2. Find \(a\) so that \(f\) is continuous at \(2\). (2 marks)
- 3. With this value of \(a\), justify that \(f\) is continuous on \(\mathbb{R}\). (2 marks)
Exercise II — IVT and bisection (7 marks)
Let \(g(x) = x^3 + x^2 - 2\).
- 1. Justify that \(g\) is continuous on \(\mathbb{R}\). (1 mark)
- 2. Compute \(g(0)\) and \(g(2)\), then show that \(g(x) = 0\) has a solution in \([0;2]\). (2 marks)
- 3. Study the sign of \(g'\) on \([0;2]\) and deduce the uniqueness of that solution \(\alpha\). (2 marks)
- 4. Bracket \(\alpha\) to within \(0{,}5\) by bisection. (2 marks)
Exercise III — Bijection and inverse (7 marks)
Let \(h(x) = \sqrt{2x + 3}\) on \(I = \left[-\tfrac{3}{2} ; +\infty\right[\).
- 1. Justify that \(h\) is continuous on \(I\). (2 marks)
- 2. Study the monotonicity of \(h\) on \(I\). (2 marks)
- 3. Determine \(J = h(I)\). (1 mark)
- 4. Show that \(h\) is a bijection from \(I\) onto \(J\) and give \(h^{-1}\) explicitly. (2 marks)
Time management advice. About \(15\) minutes for Exercise I, \(20\) for II, \(20\) for III, and \(5\) minutes of proofreading. On every question that invokes a theorem, check that you have written down its hypotheses.
Paper 2 — Limits and continuity
\(1\)-hour paper, \(20\) marks. No documents allowed. Answer the three exercises in any order, writing out your reasoning in full.
Exercise I — Computing limits (6 marks)
1. Compute \(\displaystyle\lim_{x \to 1} \frac{x^2 - 1}{x^2 - 3x + 2}\). (2 marks)
2. Compute \(\displaystyle\lim_{x \to +\infty}\left(\sqrt{x^2+4x} - x\right)\). (2 marks)
3. Compute \(\displaystyle\lim_{x \to 0} \frac{\sin(5x)}{2x}\). (2 marks)
Exercise II — Squeeze and comparison (7 marks)
Let \(f\) be defined on \(]0;+\infty[\) by \(f(x) = \dfrac{2 + \cos x}{x}\).
1. Show that \(\dfrac{1}{x} \leqslant f(x) \leqslant \dfrac{3}{x}\) for every \(x > 0\). (3 marks)
2. Deduce \(\displaystyle\lim_{x \to +\infty} f(x)\). (2 marks)
3. Let \(g(x) = x + \sin x\). Show that \(\displaystyle\lim_{x \to +\infty} g(x) = +\infty\). (2 marks)
Exercise III — Extension by continuity (7 marks)
Let \(h(x) = \dfrac{x^2 - 3x + 2}{x - 1}\) defined on \(]1;+\infty[\).
1. Determine \(\displaystyle\lim_{x \to 1^{+}} h(x)\). (2 marks)
2. Can \(h\) be extended by continuity at \(1\)? If so, give its extension \(\tilde h\). (3 marks)
3. Let \(k(x) = \sqrt{h(x) + 3}\). Compute \(\displaystyle\lim_{x \to 1^{+}} k(x)\), justifying the passage to the limit. (2 marks)
Exam mode — solutions
Exercise I (6 marks)
1. For \(x > 2\), factorise the numerator. \(x^2 - x - 2\) has roots \(2\) and \(-1\), hence
\[ \frac{x^2-x-2}{x-2} = \frac{(x-2)(x+1)}{x-2} = x+1 \] \[ \lim_{x \to 2^+} f(x) = 3 \]2. On the left, \(\displaystyle\lim_{x \to 2^-} f(x) = 2a + 1 = f(2)\). Continuity at \(2\) forces
\[ 2a + 1 = 3 \;\Longrightarrow\; a = 1 \]3. With \(a = 1\): on \(]-\infty;2]\), \(f(x) = x+1\) is a polynomial hence continuous; on \(]2;+\infty[\), \(f(x) = x+1\) as well. At \(2\), continuity was established in question 2. So \(f\) is continuous on \(\mathbb{R}\) — and in fact equals \(x+1\) everywhere.
Exercise II (7 marks)
1. \(g\) is a polynomial, hence continuous on \(\mathbb{R}\).
2.
\[ g(0) = -2 < 0, \qquad g(2) = 8 + 4 - 2 = 10 > 0 \]\(g\) is continuous on \([0;2]\) and \(g(0) \cdot g(2) < 0\): by the IVT, \(g\) vanishes at least once in \(]0;2[\).
3.
\[ g'(x) = 3x^2 + 2x = x(3x+2) \]On \(]0;2]\), \(x > 0\) and \(3x + 2 > 0\) so \(g' > 0\): \(g\) is strictly increasing on \([0;2]\). The solution \(\alpha\) is therefore unique.
4. Bisection:
\[ g(1) = 1 + 1 - 2 = 0 \]The solution is reached exactly: \(\alpha = 1\). The required bracketing is therefore immediate, \(\alpha \in [1;1]\).
Remark. A candidate who continues the bisection without noticing \(g(1) = 0\) is not penalised if the bracketing obtained is correct.
Exercise III (7 marks)
1. Set \(u(x) = 2x+3\), a polynomial hence continuous on \(I\), and \(v(t) = \sqrt{t}\), continuous on \([0;+\infty[\). For \(x \geqslant -\tfrac{3}{2}\), \(u(x) \geqslant 0\), so \(u(I) \subset [0;+\infty[\). By composition, \(h = v \circ u\) is continuous on \(I\).
2. \(u\) is strictly increasing and \(v\) is strictly increasing on \([0;+\infty[\): the composite \(h\) is strictly increasing on \(I\).
Or by computation, on \(\left]-\tfrac32;+\infty\right[\):
\[ h'(x) = \frac{2}{2\sqrt{2x+3}} = \frac{1}{\sqrt{2x+3}} > 0 \]3.
\[ h\!\left(-\tfrac{3}{2}\right) = 0, \qquad \lim_{x \to +\infty} h(x) = +\infty \]\(h\) continuous and strictly increasing on \(I\) gives \(J = h(I) = [0;+\infty[\).
4. Continuous and strictly monotonic: \(h\) is a bijection from \(I\) onto \(J\). For \(y \geqslant 0\):
\[ y = \sqrt{2x+3} \;\Longrightarrow\; y^2 = 2x+3 \;\Longrightarrow\; x = \frac{y^2-3}{2} \] \[ h^{-1}(y) = \frac{y^2 - 3}{2} \quad \text{for } y \in [0;+\infty[ \]Check. \(h(3) = \sqrt{9} = 3\) and \(h^{-1}(3) = \frac{9-3}{2} = 3\) ✓
How to read your score
- 16 to 20 — Chapter mastered. Move on to the next chapter.
- 11 to 15 — Foundations in place. Redo the level 3 exercises.
- 6 to 10 — Review the Core course Part C, then worked examples 4 and 5.
- 0 to 5 — Start the chapter again from Part B, and redo the diagnostic in the Welcome section.
Most common mistakes
- Exercise I.3: forgetting to treat both pieces and the matching point.
- Exercise II.3: concluding uniqueness without studying the sign of \(g'\).
- Exercise III.1: invoking composition without checking \(u(I) \subset [0;+\infty[\).
- Exercise III.4: giving \(h^{-1}\) without stating its domain of definition.
Paper 2 — solutions
Exercise I (6 marks)
1. The number \(1\) cancels both polynomials:
\[ \frac{x^2-1}{x^2-3x+2} = \frac{(x-1)(x+1)}{(x-1)(x-2)} = \frac{x+1}{x-2} \quad (x \neq 1) \] \[ \lim_{x \to 1} \frac{x^2-1}{x^2-3x+2} = \frac{2}{-1} = -2 \]2. Conjugate expression:
\[ \sqrt{x^2+4x} - x = \frac{4x}{\sqrt{x^2+4x}+x} = \frac{4}{\sqrt{1+\frac{4}{x}}+1} \longrightarrow 2 \]3. Bring out \(\dfrac{\sin t}{t}\):
\[ \frac{\sin(5x)}{2x} = \frac{5}{2} \cdot \frac{\sin(5x)}{5x} \longrightarrow \frac{5}{2} \]Exercise II (7 marks)
1. For every real \(x\), \(-1 \leqslant \cos x \leqslant 1\), hence \(1 \leqslant 2+\cos x \leqslant 3\). Dividing by \(x > 0\), which does not reverse the inequalities:
\[ \frac{1}{x} \leqslant \frac{2+\cos x}{x} \leqslant \frac{3}{x} \]2. Both bounds tend to \(0\) at \(+\infty\). By the squeeze theorem:
\[ \lim_{x \to +\infty} f(x) = 0 \]3. Since \(\sin x \geqslant -1\), we have \(g(x) \geqslant x - 1\) for every \(x\). And \(x-1 \to +\infty\). By comparison:
\[ \lim_{x \to +\infty} g(x) = +\infty \]A single lower bound is enough here, because we are pushing towards \(+\infty\).
Exercise III (7 marks)
1. The numerator \(x^2-3x+2 = (x-1)(x-2)\) also vanishes at \(1\). After simplification, valid since \(x \neq 1\):
\[ h(x) = x - 2 \qquad \text{so} \qquad \lim_{x \to 1^{+}} h(x) = -1 \]The limit is finite: the denominator vanishes, yet there is no divergence, contrary to what the shape of the expression suggested.
2. The limit exists and is finite: \(h\) can be extended by continuity at \(1\), and the extension is unique:
\[ \tilde h(x) = \begin{cases} h(x) & \text{if } x > 1 \\ -1 & \text{if } x = 1 \end{cases} \]The value \(-1\) is not chosen: it is imposed by the limit.
3. We have \(h(x) + 3 \to -1 + 3 = 2\) as \(x \to 1^{+}\). The function \(t \mapsto \sqrt{t}\) is continuous at \(2\): the passage to the limit in the composite is valid with no extra hypothesis.
\[ \lim_{x \to 1^{+}} k(x) = \sqrt{2} \]How to read your score — Paper 2
16 to 20 — limits are secure, including the tricky cases.
11 to 15 — the techniques are known, the justifications still need care.
6 to 10 — review the indeterminate forms and the squeeze theorem before going further.
0 to 5 — go back to Part A of the course — limits — and to examples 6 to 9.
Most common mistakes — Paper 2
Concluding divergence as soon as the denominator vanishes, without checking whether the numerator vanishes too — Exercise III question 1.
Dividing an inequality chain by \(x\) without stating that \(x > 0\): with \(x < 0\), the inequalities would reverse.
Using the squeeze theorem to conclude an infinite limit: the squeeze theorem gives a finite limit; for infinity it is comparison, and a single bound suffices.