IB Analysis and Approaches HL · Differential calculus · Section 11 of 14

Rolle's and mean value theorems; L'Hôpital's rule

3 worked examples · By Álvaro López

Key idea

Two theorems guarantee that, if a function is well behaved on an interval, there is a point where the tangent has a particular gradient. L'Hôpital's rule uses this idea to evaluate limits that give $\frac{0}{0}$ or $\frac{\infty}{\infty}$ by differentiating numerator and denominator. This section is Higher Level content.

Rolle's theorem

The idea: a curve that leaves from one height and returns to the same height must rise and fall, or fall and rise. At some intermediate point it turns around, and there the tangent is horizontal. Geometrically: there is a point on the graph where the tangent is parallel to the $x$ -axis.

A useful consequence: between two zeros of $f$ there is at least one zero of $f^{'}$ , provided $f$ satisfies the hypotheses.

Mean value theorem

The fraction on the right is the gradient of the chord joining $(a, f (a))$ and $(b, f (b))$ , that is, the average rate of change. The theorem says that at some point the instantaneous gradient equals it: there is a tangent parallel to the chord. Rolle's theorem is the special case $f (a) = f (b)$ , in which the chord is horizontal.

Graph of the square root on the interval (1, 9): the tangent at x=4 is parallel to the chord joining the endpoints

L'Hôpital's rule

When a limit of a quotient gives $\frac{0}{0}$ or $\frac{\infty}{\infty}$ , you cannot know in advance what it equals. L'Hôpital's rule resolves it by replacing the quotient with the quotient of the derivatives.

The rule rests on the mean value theorem: near $a$ , the quotient $f / g$ behaves like the quotient of the gradients of the tangents, that is, like $f^{'} / g^{'}$ .

Other indeterminate forms

The rule only applies to $\frac{0}{0}$ and $\frac{\infty}{\infty}$ , but other forms can be converted into one of them:

$0 \cdot \infty$ : write the product as a quotient, $f \cdot g = \frac{f}{1/ g}$ .
$\infty - \infty$ : combine the fractions into one, or factor out a common term.
$1^{\infty}$ , $0^{0}$ , $\infty^{0}$ : if $L = lim f^{g}$ , take logarithms: $ln L = lim g ln f$ , which is of the form $0 \cdot \infty$ . Then $L = e^{l n L}$ .

Worked examples

Example 1 Basic

Rolle's theorem for a cubic

Check that $f (x) = x^{3} - 4 x$ satisfies the hypotheses of Rolle's theorem on $[- 2, 2]$ , and find the values of $c$ given by the theorem.

Solution

$f$ is a polynomial: it is continuous and differentiable on all of $R$ . Also $f (- 2) = - 8 + 8 = 0$ and $f (2) = 8 - 8 = 0$ , so $f (- 2) = f (2)$ .

By Rolle's theorem there exists $c \in] - 2, 2 [$ with $f^{'} (c) = 0$ . $f^{'} (x) = 3 x^{2} - 4 = 0 \Rightarrow x = \pm \frac{2}{3} = \pm \frac{2 3}{3} \approx \pm 1.15$ .

Both values lie in $] - 2, 2 [$ : $c = \frac{2 3}{3}$ and $c = - \frac{2 3}{3}$ .

Example 2 Medium

Mean value theorem and a tangent parallel to the chord

Let $f (x) = x + 1$ on the interval $[0, 8]$ .

a) Find the value of $c \in] 0, 8 [$ given by the mean value theorem.

b) Write the equation of the tangent to the graph at $x = c$ , in the form $y = m x + n$ .

Solution

a) $f$ is continuous on $[0, 8]$ and differentiable on $] 0, 8 [$ . The gradient of the chord is $\frac{f ( 8 ) - f ( 0 )}{8 - 0} = \frac{3 - 1}{8} = \frac{1}{4} .$ Solve $f^{'} (c) = \frac{1}{4}$ , with $f^{'} (x) = \frac{1}{2 x + 1}$ : $\frac{1}{2 c + 1} = \frac{1}{4} \Rightarrow c + 1 = 2 \Rightarrow c = 3.$ It lies in $] 0, 8 [$ .

b) $f (3) = 2$ and the gradient is $\frac{1}{4}$ : $y - 2 = \frac{1}{4} (x - 3)$ , that is, $y = \frac{x}{4} + \frac{5}{4}$ .

Example 3 Hard

Three limits using L'Hôpital

Calculate, without a calculator:

a) $x \to 0 lim \frac{e ^{2 x} - 1 - 2 x}{x ^{2}}$

b) $x \to 0^{+} lim x ln x$

c) $x \to + \infty lim (1 + \frac{3}{x})^{x}$

Solution

a) Substituting $x = 0$ gives $\frac{0}{0}$ . Differentiating numerator and denominator: $x \to 0 lim \frac{2 e ^{2 x} - 2}{2 x} (\frac{0}{0}) .$ It is still $\frac{0}{0}$ ; differentiate again: $x \to 0 lim \frac{4 e ^{2 x}}{2} = 2.$

b) It is of the form $0 \cdot (- \infty)$ . Write it as a quotient: $x ln x = \frac{ln x}{1/ x}$ , which gives $\frac{- \infty}{+ \infty}$ . Applying the rule: $x \to 0^{+} lim \frac{1/ x}{- 1/ x ^{2}} = x \to 0^{+} lim (- x) = 0.$

c) It is of the form $1^{\infty}$ . Let $L = x \to \infty lim (1 + \frac{3}{x})^{x}$ . Then $ln L = x \to \infty lim x ln (1 + \frac{3}{x}) = x \to \infty lim \frac{ln ( 1 + 3/ x )}{1/ x} (\frac{0}{0}) .$ Differentiating: the numerator gives $\frac{1}{1 + 3/ x} \cdot (- \frac{3}{x ^{2}})$ and the denominator $- \frac{1}{x ^{2}}$ . The quotient is $\frac{3}{1 + 3/ x} \to 3$ . So $ln L = 3$ and $L = e^{3}$ .

In the exam: state the indeterminate form before each application of the rule and write $lim$ in every equality. Applying the rule without checking that you have an indeterminate form, or writing the final value without the steps, loses method marks.

Common mistakes

Summary

Rolle: $f$ continuous on $[a, b]$ , differentiable on $] a, b [$ and $f (a) = f (b)$ $\Rightarrow$ there exists $c$ with $f^{'} (c) = 0$ .
Mean value theorem: there exists $c \in] a, b [$ with $f^{'} (c) = \frac{f ( b ) - f ( a )}{b - a}$ : a tangent parallel to the chord.
L'Hôpital: if $\frac{f}{g}$ gives $\frac{0}{0}$ or $\frac{\infty}{\infty}$ , then $lim \frac{f}{g} = lim \frac{f ^{'}}{g ^{'}}$ (differentiate each part separately).
$0 \cdot \infty$ and $\infty - \infty$ are turned into a quotient; $1^{\infty}$ , $0^{0}$ and $\infty^{0}$ are handled with logarithms.
Always check the hypotheses and the indeterminate form before applying.

Keep studying

Maths classes in Aldaia and online

Want to work on it together?

Free first consultation: we look at your level and your exam, and build a plan. Álvaro replies on WhatsApp.

Message us on WhatsApp