proof of l’Hôpital’s rule for $\mathrm{\infty }/\mathrm{\infty }$ form

This is the proof of L’Hôpital’s Rule (http://planetmath.org/LHpitalsRule)in the case of the indeterminate form $\pm \mathrm{\infty }/\mathrm{\infty }$.Compared to the proof for the $0/0$ case (http://planetmath.org/ProofOfDeLHopitalsRule),more complicated estimates are needed.

Assume that

$$\underset{x\to a}{lim}f(x)=\pm \mathrm{\infty },\underset{x\to a}{lim}g(x)=\pm \mathrm{\infty },\underset{x\to a}{lim}\frac{f^{\prime }(x)}{g^{\prime }(x)}=m,$$

where $a$ and $m$ are real numbers.The case when $a$ or $m$ is infinite only involvesslight modifications to the arguments below.

Given $ϵ>0$.there is a $\delta >0$ such that

whenever .

Let $c$ and $x$ be points such that or .(That is, both $c$ and $x$ are within distance $\delta$ of $a$,but $x$ is always closer.)By Cauchy’s mean value theorem, there exists some ${\xi }_x$in between $c$ and $x$ (and hence )such that

$$\frac{f(x)-f(c)}{g(x)-g(c)}=\frac{f^{\prime }({\xi }_x)}{g^{\prime }({\xi }_x)}.$$

We can assume the values $f(x)$, $g(x)$, $f(x)-f(c)$, $g(x)-g(c)$are all non-zero when $x$ is close enough to $a$,say, when for some .(So there is no division by zero in our equations.)This is because $f(x)$ and $g(x)$ were assumed to approach $\pm \mathrm{\infty }$,so when $x$ is close enough to $a$,they will exceed the fixed values $f(c)$, $g(c)$, and $0$.

We write

${\displaystyle \frac{f(x)}{g(x)}}$$={\displaystyle \frac{f(x)}{f(x)-f(c)}}\cdot {\displaystyle \frac{g(x)-g(c)}{g(x)}}\cdot {\displaystyle \frac{f(x)-f(c)}{g(x)-g(c)}}$

$={\displaystyle \frac{1-g(c)/g(x)}{1-f(c)/f(x)}}\cdot {\displaystyle \frac{f^{\prime }({\xi }_x)}{g^{\prime }({\xi }_x)}}.$

Note that

$$\underset{x\to a}{lim}\frac{1-g(c)/g(x)}{1-f(c)/f(x)}=1,$$

but ${\xi }_x$ is not guaranteed to approach $a$ as $x$ approaches $a$,so we cannot just take the limit $x\to a$ directly. However:there exists so that

whenever .Then

$\left|{\displaystyle \frac{f(x)}{g(x)}}-m\right|$$=\left|\left({\displaystyle \frac{f^{\prime }({\xi }_x)}{g^{\prime }({\xi }_x)}}-m\right)+{\displaystyle \frac{f^{\prime }({\xi }_x)}{g^{\prime }({\xi }_x)}}\left({\displaystyle \frac{1-g(c)/g(x)}{1-f(c)/f(x)}}-1\right)\right|$

$\le ϵ+(|m|+ϵ){\displaystyle \frac{ϵ}{|m|+ϵ}}=2ϵ$

for .

This proves

$$\underset{x\to a}{lim}\frac{f(x)}{g(x)}=m=\underset{x\to a}{lim}\frac{f^{\prime }(x)}{g^{\prime }(x)}.$$