Differentiable functions/Mean value theorem/General/L'Hôpital's rule/Section

The following statement is called also the general mean value theorem.

Theorem

Let ${}b>a$, and suppose that

$f,g\colon [a,b]\longrightarrow \mathbb {R}$

are continuous functions which are differentiable on ${}]a,b[$ and such that

${}g'(x)\neq 0\,$

for all ${}x\in {]a,b[}$. Then ${}g(b)\neq g(a)$, and there exists some ${}c\in {]a,b[}$ such that

${}{\frac {f(b)-f(a)}{g(b)-g(a)}}={\frac {f'(c)}{g'(c)}}\,.$

Proof

The statement

${}g(a)\neq g(b)\,$

follows from fact. We consider the auxiliary function

${}h(x):=f(x)-{\frac {f(b)-f(a)}{g(b)-g(a)}}g(x)\,.$

We have

${}{\begin{aligned}h(a)&=f(a)-{\frac {f(b)-f(a)}{g(b)-g(a)}}g(a)\\&={\frac {f(a)g(b)-f(a)g(a)-f(b)g(a)+f(a)g(a)}{g(b)-g(a)}}\\&={\frac {f(a)g(b)-f(b)g(a)}{g(b)-g(a)}}\\&={\frac {f(b)g(b)-f(b)g(a)-f(b)g(b)+f(a)g(b)}{g(b)-g(a)}}\\&=f(b)-{\frac {f(b)-f(a)}{g(b)-g(a)}}g(b)\\&=h(b).\end{aligned}}$

Therefore, ${}h(a)=h(b)$, and fact yields the existence of some ${}c\in {]a,b[}$ with

${}h'(c)=0\,.$

Rearranging proves the claim.

$\Box$

From this version, one can recover the mean value theorem, by taking for ${}g$ the identity.

L’Hospital (1661-1704)

For the computation of the limit of a function, the following method called L'Hôpital's rule helps.

Corollary

Let ${}I\subseteq \mathbb {R}$ denote an open interval, and let ${}a\in I$ denote a point. Suppose that

$f,g\colon I\longrightarrow \mathbb {R}$

are continuous functions, which are differentiable on ${}I\setminus \{a\}$, fulfilling ${}f(a)=g(a)=0$, and with ${}g'(x)\neq 0$ for ${}x\neq a$. Moreover, suppose that the limit

${}w:=\operatorname {lim} _{x\rightarrow a}\,{\frac {f'(x)}{g'(x)}}\,$

exists. Then also the limit

$\operatorname {lim} _{x\rightarrow a}\,{\frac {f(x)}{g(x)}}$

exists, and it also equals ${}w$.

Proof

Because ${}g'$ has no zero in the interval and ${}g(a)=0$ holds, it follows, because of fact, that ${}a$ is the only zero of ${}g$. Let ${}{\left(x_{n}\right)}_{n\in \mathbb {N} }$ denote a sequence in ${}I\setminus \{a\}$, converging to ${}a$.

For every ${}x_{n}$ there exists, due to fact, applied to the interval ${}I_{n}:=[x_{n},a]$ or ${}[a,x_{n}]$, a ${}c_{n}$ (in the interior of ${}I_{n}$,) fulfilling

${}{\frac {f(x_{n})-f(a)}{g(x_{n})-g(a)}}={\frac {f'(c_{n})}{g'(c_{n})}}\,.$

The sequence ${}{\left(c_{n}\right)}_{n\in \mathbb {N} }$ converges also to ${}a$, so that, because of the condition, the right-hand side converges to ${}{\frac {f'(a)}{g'(a)}}=w$. Therefore, also the left-hand side converges to ${}w$, and, because of ${}f(a)=g(a)=0$, this means that ${}{\frac {f(x_{n})}{g(x_{n})}}$ converges to ${}w$.

$\Box$

Example

The polynomials

$3x^{2}-5x-2{\text{ and }}x^{3}-4x^{2}+x+6$

have both a zero for ${}x=2$. It is therefore not immediately clear whether the limit

$\operatorname {lim} _{x\rightarrow 2}\,{\frac {3x^{2}-5x-2}{x^{3}-4x^{2}+x+6}}$

exists. Applying twice L'Hôpital's rule, we get the existence and

${}\operatorname {lim} _{x\rightarrow 2}\,{\frac {3x^{2}-5x-2}{x^{3}-4x^{2}+x+6}}=\operatorname {lim} _{x\rightarrow 2}\,{\frac {6x-5}{3x^{2}-8x+1}}={\frac {7}{-3}}=-{\frac {7}{3}}\,.$