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

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

Theorem

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

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

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

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

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

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

Proof

The statement

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

follows from fact. We consider the auxiliary function

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

We have

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

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

Rearranging proves the claim.

${\displaystyle \Box }$

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

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

Corollary

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

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

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

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

exists. Then also the limit

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

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

Proof

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

For every ${\displaystyle {}x_{n}}$ there exists, due to fact, applied to the interval ${\displaystyle {}I_{n}:=[x_{n},a]}$ or ${\displaystyle {}[a,x_{n}]}$, a ${\displaystyle {}c_{n}}$ (in the interior[1] of ${\displaystyle {}I_{n}}$,) fulfilling

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

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

${\displaystyle \Box }$

Example

The polynomials

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

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

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

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

${\displaystyle {}\operatorname {lim} _{x\rightarrow 2}\,{\frac {3x^{2}-5x-2}{x^{3}-4x^{2}+x+6}}=\operatorname {lim} _{x\rightarrow 2}\,{\frac {6x-5}{3x^{2}-8x+1}}={\frac {7}{-3}}=-{\frac {7}{3}}\,.}$
1. The interior of a real interval ${\displaystyle {}I\subseteq \mathbb {R} }$

is the interval without the boundaries.