Is the following a valid proof of the $\frac{0}{0}$ case of L'Hôpital's rule?

Asked Jan 02 '18 at 16:18

Active Jan 02 '18 at 17:30

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Let $\displaystyle \lim_{x\to c}~f(x)=\lim_{x\to c}~g(x)=0$ and $g'(c)\neq0$, Edit: Assume also that $c$ lies in the open interval $S$, and that $f(x)$ and $g(x)$ is continuous on $S$, Then consider $\displaystyle \lim_{x\to c}\frac{f(x)}{g(x)}$, $$=\lim_{x\to c}\frac{f(x)-0}{g(x)-0}$$ $$=\lim_{x\to c}\frac{f(x)-f(c)}{g(x)-g(c)}$$ $$=\lim_{x\to c}\frac{\left[\frac{f(x)-f(c)}{x-c}\right]}{\left[\frac{g(x)-g(c)}{x-c}\right]}$$ $$=\frac{\lim_{x\to c}\left[\frac{f(x)-f(c)}{x-c}\right]}{\lim_{x\to c}\left[\frac{g(x)-g(c)}{x-c}\right]}$$ $$=\frac{f'(c)}{g'(c)}$$ $$=\lim_{x\to c}\frac{f'(x)}{g'(x)}$$hence, when $\displaystyle \lim_{x\to c}\frac{f(x)}{g(x)}=\frac{0}{0}$, then $\displaystyle \lim_{x\to c}\frac{f(x)}{g(x)}=\lim_{x\to c}\frac{f'(x)}{g'(x)}=L$

Is this correct, or have I missed anything? Also, can someone aid me in proving the $\frac{\infty}{\infty}$ case? Any responses are appreciated.

edited Jan 02 '18 at 17:30

asked Jan 02 '18 at 16:18

joshuaheckroodt

2,789

You are assuming that $f'/g'$ is continuous at $c$. – Angina Seng Jan 02 '18 at 16:22
I've added it now, thank you! @LordSharktheUnknown – joshuaheckroodt Jan 02 '18 at 16:24
1

You have included as a hypothesis that $f'/g'$ is continuous on the given interval. Standard treatments, for instance Rudin's Principles, neither make nor use such an assumption. – Angina Seng Jan 02 '18 at 16:33
For $\frac{\infty}{\infty}$ case do the same with $\frac {\frac1g}{\frac1f}$ – QED Jan 02 '18 at 16:39
You just need the assumption that $f$ and $g$ are differentiable on the interval in question (this does not imply that the functions' derivative is continuous). – user351797 Jan 02 '18 at 16:42
@AbishankaSaha using that method one can't prove the general L'Hôpital's rule – ℋolo Jan 02 '18 at 17:01

Is the following a valid proof of the $\frac{0}{0}$ case of L'Hôpital's rule?

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