Limit of the reciprocal without proving its existence

Question

Suppose $\lim_{x\to c}f(x)=\infty$ and $\lim_{x\to c}g(x)=\infty$. Also, $\frac{d}{dx}f(x)=g(x)$ and $\frac{d}{dx}g(x)=f(x)$. If I apply L'Hospital's rule: $$L=\lim_{x\to c}\frac{f(x)}{g(x)}=\lim_{x\to c}\frac{g(x)}{f(x)}=\frac{1}{L}$$ Is this sufficient to say that $L=1$ or do I have to prove that both limits exist?

Answer 1

You will indeed have to prove that the limit exists, L'Hôpital's rule rule only says that $\lim_{x\rightarrow c}\frac{f(x)}{g(x)}=\lim_{x\rightarrow c}\frac{f'(x)}{g'(x)}$, iff the second limit exists.

Answer 2

If $f'=g$ and $g'=f$, then $f''=f$ so $$f(x) = Ae^x + Be^{-x}$$ For constants $A,B$ are the only functions verifying your hypothesis. So you're actually looking at the limit

$$\lim_{x\to c} \frac{Ae^x + Be^{-x}}{Ae^x - Be^{-x}}$$

If $c$ is finite, the value is 1 only if $B=0$.