Rolle, Lagrange and L'Hôpital theorems: worked exercises

In Year 5 of high school, derivatives start between mid-October and November. The theorems (Rolle, Lagrange, Cauchy, L’Hôpital) usually arrive at the end of November, when the differentiation rules have already been explained. The test on derivatives falls in December, that is eight to ten weeks from today. This guide applies to the Italian liceo scientifico (science-focused high school) and to every high school that has maths in Year 5. I am publishing it now so that it is already readable when the teacher reaches the theorems: in my lessons I see that this is the point where many students stop understanding and start copying.

Here I talk only about the theorems. For the differentiation rules and the points of non-differentiability there is derivatives explained properly, and for the surrounding calculus you will find limits and asymptotes and function study with the first derivative. The picture of the whole year is in the Year 5 programme and formula sheet.

The method in one line

A theorem of calculus is a machine: if the hypotheses are true, the conclusion follows. If even one hypothesis is missing, the machine says nothing (it doesn’t say the conclusion is false, it only says you can’t use it). The test gives marks for two distinct things: having checked the hypotheses and having drawn the conclusion. Whoever skips the check loses half the marks even if the result is right.

Rolle’s theorem

Statement. Let $f$ be a function such that:

1. $f$ is continuous on the closed interval $[a,b]$;
2. $f$ is differentiable on the open interval $(a,b)$;
3. $f(a)=f(b)$.

Then there exists at least one point $c\in (a,b)$ such that

Geometric meaning. If the graph starts and ends at the same height, somewhere in between there is a horizontal tangent. The point can be a maximum, a minimum or an inflection point with a horizontal tangent.

Mistake number 1: saying “Rolle applies” without checking. Take $f(x)=|x|$ on $[-1,1]$. The function is continuous, and $f(-1)=f(1)=1$. But there is no point with zero derivative: to the left of zero the derivative is $-1$, to the right it is $+1$, and at $x=0$ the function is not differentiable. Here the second hypothesis fails. The theorem is not wrong: it simply does not apply.

Lagrange’s theorem (the mean value theorem)

Statement. Let $f$ be continuous on $[a,b]$ and differentiable on $(a,b)$. Then there exists $c\in (a,b)$ such that

Geometric meaning. There is a point where the tangent is parallel to the chord joining the endpoints of the graph. Rolle is the special case in which the chord is horizontal, that is $f(a)=f(b)$.

Physical meaning. If you cover 150 km in two hours, the average speed is 75 km/h, and at least at one instant the speedometer read exactly 75. It is the argument by which an authority can fine someone without ever having seen them exceed the limit: it is enough to compare two checkpoints.

Two corollaries that often come up in tests.

• If $f^{\prime }(x)=0$ on a whole interval, then $f$ is constant on that interval. As a consequence two functions with the same derivative differ by a constant: that is why the "$+c$" appears in the integral.
• If $f^{\prime }(x)>0$ on an interval, then $f$ is strictly increasing (if $f^{\prime }(x)<0$, decreasing). It is the theoretical justification of the method of studying the sign of the first derivative.

Cauchy in three lines

Given $f$ and $g$ continuous on $[a,b]$ and differentiable on $(a,b)$, with $g^{\prime }(x)\ne 0$ on $(a,b)$, there exists $c\in (a,b)$ such that

With $g(x)=x$ you get Lagrange back. It serves above all as a step for proving L’Hôpital: in an oral exam it may be asked for, in a written test almost never.

How to check applicability

When the question says “check that the function satisfies the hypotheses of the … theorem on the interval … and determine the points $c$”, the answer has three parts, always in the same order.

1. Continuity on $[a,b]$. An elementary function (polynomial, exponential, sine, cosine) is continuous where it is defined. For rational functions, roots and logarithms, look at where the denominator vanishes or where the argument leaves the domain. For piecewise-defined functions compute the right-hand and left-hand limits at the junction point and compare them with the value.
2. Differentiability on $(a,b)$. Compute the derivative and find where it doesn’t exist. Absolute values, roots with fractional exponents and piecewise junctions are the usual suspects. Remember that the endpoints don’t count: interior points are enough.
3. For Rolle, $f(a)=f(b)$. Compute the two values and compare them, writing them down.

Only after this do you write “the hypotheses are satisfied, so $c$ exists” and move on to finding it. The point $c$ is found by solving the equation $f^{\prime }(x)=0$ (Rolle) or $f^{\prime }(x)=\frac{f(b)-f(a)}{b-a}$ (Lagrange), and then discarding the solutions outside $(a,b)$. I often see this forgotten: the equation has two solutions, only one is in the interval, and the student lists both.

Exercise 1: checking Rolle and finding $c$

Check that $f(x)=x^3-3x$ satisfies the hypotheses of Rolle’s theorem on $[0,\sqrt{3}]$ and determine the points $c$.

Continuity and differentiability. $f$ is a polynomial, so it is continuous and differentiable on all of $\mathbb{R}$, in particular on $[0,\sqrt{3}]$ and on $(0,\sqrt{3})$.

Values at the endpoints.

So $f(0)=f(\sqrt{3})$ and the three hypotheses are verified.

Finding $c$. The derivative is $f^{\prime }(x)=3x^2-3$. We set $f^{\prime }(x)=0$:

Only $x=1$ belongs to $(0,\sqrt{3})$, because $\sqrt{3}\approx 1.73$. So $c=1$. The value $x=-1$ must be discarded: it is outside the interval (it would be another valid point if the interval were $[-\sqrt{3},\sqrt{3}]$).

Before moving on to the next exercises, if you want to train on checking the hypotheses on many cases, there is the Year 5 maths test on derivatives: the correct answer is free, the guided solution unlocks with the first lesson.

Exercise 2: Lagrange and an inequality

Part a. Find the point $c$ of Lagrange’s theorem for $f(x)=x^3-2x$ on $[0,2]$.

$f$ is a polynomial, so continuous and differentiable everywhere. We compute the values at the endpoints:

The slope of the chord is

The derivative is $f^{\prime }(x)=3x^2-2$. We set $f^{\prime }(x)=2$:

Only the positive value lies in $(0,2)$, because $\frac{2}{\sqrt{3}}\approx 1.15$. So $c=\frac{2}{\sqrt{3}}=\frac{2\sqrt{3}}{3}$.

Part b. Prove that for all real $a,b$ we have

If $a=b$ the inequality is $0\le 0$ and it is true. Suppose $a<b$ (the other case is symmetric, because both sides are unchanged when $a$ and $b$ are swapped). The function $\sin x$ is continuous and differentiable on all of $\mathbb{R}$, so on $[a,b]$ Lagrange applies: there exists $c\in (a,b)$ with

that is $\sin b-\sin a=\cos c(b-a)$. Taking absolute values:

because $|\cos c|\le 1$ for every $c$. The technique is always the same: write the difference quotient, replace it with the derivative evaluated at an unknown point, then bound the derivative above or below with a known estimate.

L’Hôpital: conditions and when not to use it

Statement. Let $f$ and $g$ be differentiable in a neighbourhood of $x_0$ (excluding at most $x_0$), with $g^{\prime }(x)\ne 0$ in that neighbourhood. If the limit of the ratio $\frac{f(x)}{g(x)}$ as $x\to x_0$ presents itself in the form $\frac{0}{0}$ or $\frac{\infty }{\infty }$, and if the limit exists

then the original limit exists and has the same value. The theorem also holds for $x\to \pm \infty$ and for one-sided limits.

The three conditions to declare in your working: the indeterminate form, the differentiability, the existence of the limit of the ratio of the derivatives. If the latter doesn’t exist, the theorem says nothing about the original. Classic example: ${\lim }_{x\to +\infty }\frac{x+\sin x}{x}$ is of type $\frac{\infty }{\infty }$, the ratio of the derivatives is $1+\cos x$, which oscillates and has no limit. Yet the true limit exists and equals $1$ (divide numerator and denominator by $x$). L’Hôpital is not needed here and doesn’t work.

When NOT to use it.

• Forms that aren’t indeterminate. ${\lim }_{x\to 0}\frac{x+2}{x}$ is not $\frac{0}{0}$: the numerator tends to $2$. Whoever differentiates anyway gets $\frac{1}{1}=1$, a wrong answer (the limit is infinite, $+\infty$ from the right, $-\infty$ from the left). Before differentiating always write down the form.
• Circular standard limits. To prove that ${\lim }_{x\to 0}\frac{\sin x}{x}=1$ you can’t use L’Hôpital, because the derivative of the sine is obtained precisely from that limit. In a calculation exercise you can use it without problems, but in a proof of the standard limits it would be reasoning that chases its own tail.
• When another route is shorter. A ratio of polynomials with $x\to \infty$ is solved by comparing degrees.

Forms to be reduced. L’Hôpital applies only to ratios. A $0\cdot \infty$ form is rewritten as a quotient by moving one factor to the denominator (as a reciprocal): $f\cdot g=\frac{f}{1/g}$. An $\infty -\infty$ form is rewritten with a common denominator or by taking out a common factor, until you get a ratio.

Exercise 3: three limits with L’Hôpital

a) Form $\frac{0}{0}$.

Numerator: $e^0-1-0=0$. Denominator: $0$. Form $\frac{0}{0}$, differentiable functions. We differentiate:

Still $\frac{0}{0}$. We differentiate again:

The limit of the ratio of the derivatives exists, so the initial limit equals $\frac{1}{2}$.

b) Form $0\cdot \infty$.

The first factor tends to $0$, the second to $-\infty$. We write

which as $x\to 0^{+}$ is of type $\frac{-\infty }{+\infty }$, so $\frac{\infty }{\infty }$ in absolute value. We differentiate numerator and denominator:

The limit is $0$. If I had instead written $x\ln x=\frac{x}{1/\ln x}$ the calculations would have become more complicated: the choice of which factor to put in the denominator matters, and the practical rule is to keep the logarithm in the numerator and put the power in the denominator.

c) Form $\infty -\infty$.

Common denominator:

As $x\to 0$ numerator and denominator tend to $0$: form $\frac{0}{0}$. We differentiate:

Again $\frac{0}{0}$ (numerator $\cos 0-1=0$, denominator $0+0=0$). We differentiate once more:

The limit is $0$. Note the step on the denominator: the derivative of $\sin x+x\cos x$ is $\cos x+\cos x-x\sin x=2\cos x-x\sin x$.

The mistakes I see most

• Applying the theorem without writing the hypotheses. The result may be right and the marks still halved.
• Not discarding solutions outside the interval when looking for $c$.
• Using Rolle without checking that the endpoints have the same value, or swapping Rolle and Lagrange.
• Differentiating the quotient as $\frac{f^{\prime }}{g^{\prime }}$ when taking the limit of the ratio, but also when the ratio is a function to be differentiated. L’Hôpital differentiates numerator and denominator separately; the derivative of a quotient is another rule. They are two different contexts and they get confused.
• Repeating L’Hôpital without rechecking the form at every step. When it stops being indeterminate, you stop.
• Not recognising limits that can be solved more quickly, with a standard limit or a comparison of infinities.

How to use these weeks

Until the end of November it makes sense to consolidate the differentiation rules and the derivative of a composite function, which is where tests lose the most marks. When the teacher moves on to the theorems, reread the statements and redo the three exercises on this page by hand without looking. Then try the Year 5 maths test on derivatives, in season from October to May. For the general picture of the subject there is the maths page.

If the theorems don’t add up for you, or you want to understand whether the gaps are in derivatives or in limits, write to me from the contacts page: the introductory call is free (30 minutes) and I will tell you where it is best to restart.