# Implications of the existence of $\lim_{x \to a+} \frac{f'(x)}{g'(x)}$

I'm reviewing my calculus notes of a variant of L'Hospital's rule and there's one step in the proof that I found puzzling.

The theorem states:

Let $a \in \mathbb{R}, \lim_{x \to a+} f(x) = \lim_{x \to a+} g(x) = 0$ and $\lim_{x \to a+} \frac{f'(x)}{g'(x)}$ exist. Then $$\lim_{x \to a+} \frac{f(x)}{g(x)} = \lim_{x \to a+} \frac{f'(x)}{g'(x)}.$$

The first line of the proof says that the following directly follows from the assumptions: $$\exists \delta > 0 \; \forall x \in (a, a + \delta): f'(x) \in \mathbb{R} \text{ and } g'(x) \in \mathbb{R} \setminus \{0\}.$$

However, I don't see why the derivatives should be real given that $\lim_{x \to a+} \frac{f'(x)}{g'(x)}$ could also be equal to $\pm \infty.$ Could anyone elaborate on this?

• I'm more curious how one can even speak about limits for $x\to a^+$ if one of the possible values for $a$ is $+\infty$. And even if we exclude the $a=+\infty$ case, $(a,a+\delta)$ is not a meaningful interval when $a=-\infty$ ... – Henning Makholm Dec 28 '15 at 20:16
• @Henning You gotta be a bit creative. $(a,a+\delta)=(a,-1/\delta)$ when $a=-\infty$ – A.S. Dec 28 '15 at 20:23
• Exactly as A.S. says; however, to make things simpler, I've limited $a$ to be real (that's actually the first case dealt with in the proof). – John Manak Dec 28 '15 at 20:24
• Did the theorem not require $f$ and $g$ to be differentiable in a neighborhood of the limit point (i.e., differentiable for $x\in (a,a+\delta)$)? Perhaps this was tacitly assumed. – Mark Viola Dec 28 '15 at 20:30
• No it didn't. The proof states (in the one line that I copied) that differentiability follows from the assumptions. – John Manak Dec 28 '15 at 20:32

After giving it some thought during the day, I think the idea is that since limits are defined only for real functions, and since $\lim_{x \to a+} \frac{f'(x)}{g'(x)}$ exists, that implies that there exists a neighbourhood of $a$ in which both derivatives are real.