# proof of l’Hôpital’s rule for $\infty/\infty$ form

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

Assume that

 $\lim_{x\to a}f(x)=\pm\infty\,,\quad\lim_{x\to a}g(x)=\pm\infty\,,\quad\lim_{x% \to a}\frac{f^{\prime}(x)}{g^{\prime}(x)}=m\,,$

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

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

 $\left\lvert\frac{f^{\prime}(\xi)}{g^{\prime}(\xi)}-m\right\rvert<\epsilon$

whenever $0<\lvert\xi-a\rvert<\delta$.

Let $c$ and $x$ be points such that $a-\delta or $a. (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 $0<\lvert\xi_{x}-a\rvert<\delta$) 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 $0<\lvert x-a\rvert<\delta^{\prime}$ for some $0<\delta^{\prime}<\delta$. (So there is no division by zero in our equations.) This is because $f(x)$ and $g(x)$ were assumed to approach $\pm\infty$, so when $x$ is close enough to $a$, they will exceed the fixed values $f(c)$, $g(c)$, and $0$.

We write

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

Note that

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

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 $0<\delta^{\prime\prime}<\delta^{\prime}$ so that

 $\left\lvert\frac{1-g(c)/g(x)}{1-f(c)/f(x)}-1\right\rvert<\frac{\epsilon}{% \lvert m\rvert+\epsilon}$

whenever $0<\lvert x-a\rvert<\delta^{\prime\prime}$. Then

 $\displaystyle\left\lvert\frac{f(x)}{g(x)}-m\right\rvert$ $\displaystyle=\left\lvert\left(\frac{f^{\prime}(\xi_{x})}{g^{\prime}(\xi_{x})}% -m\right)+\frac{f^{\prime}(\xi_{x})}{g^{\prime}(\xi_{x})}\left(\frac{1-g(c)/g(% x)}{1-f(c)/f(x)}-1\right)\right\rvert$ $\displaystyle\leq\epsilon+(\lvert m\rvert+\epsilon)\frac{\epsilon}{\lvert m% \rvert+\epsilon}=2\epsilon$

for $0<\lvert x-a\rvert<\delta^{\prime\prime}$.

This proves

 $\lim_{x\to a}\frac{f(x)}{g(x)}=m=\lim_{x\to a}\frac{f^{\prime}(x)}{g^{\prime}(% x)}\,.$

## References

• 1 Michael Spivak, Calculus, 3rd ed. Publish or Perish, 1994.
Title proof of l’Hôpital’s rule for $\infty/\infty$ form ProofOfLHopitalsRuleForinftyinftyForm 2013-03-22 15:40:15 2013-03-22 15:40:15 stevecheng (10074) stevecheng (10074) 7 stevecheng (10074) Proof msc 26A06