# Satisfying Hypotheses For L'Hospital's Rule

I have that the following hypothesis need to be satisfied in order to use L'Hospital's rule for, $$$$\lim_{x\to a}\frac{f(x)}{g(x)}$$$$

1. $$f(x)$$ and $$g(x)$$ have to be functions differentiable near $$a$$.

2. $$\lim_{x\to a}\frac{f(x)}{g(x)}$$ has to have the indeterminate form $$\frac{0}{0}$$ or $$\frac{\pm\infty}{\pm\infty}$$.

3. $$\lim_{x\to a}\frac{f'(x)}{g'(x)}$$ has to exist.

If possible, find the following limit using L’Hospital’s rule. If not possible, explain why, $$$$\lim_{x\to0^+}\frac{\ln{x}}{\frac{1}{\sin{x}}}$$$$ How do I check whether the hypothesis are satisfied and whether I can use L’Hospital’s rule for this or not.

I tried differentiating $$\ln(x)$$ and $$\frac{1}{\sin(x)}$$ at $$0$$ and saw that the derivative of $$\ln(x)$$ does not exist at $$0$$. From this, I came to the conclusion that L'Hospital's rule cannot be used. I tried using L'Hospital's anyway and still arrived at the correct answer, $$0$$, which I verified using Desmos. What am I doing wrong?

• The assumptions for L'Hopital's rule do not require the function to be differentiable at $x=a$, but rather on an open interval containing $a$ except possibly at $x=a$. So L'Hopital's rule is valid here. Nov 30, 2020 at 20:35
• Whether the derivative of $\ln (x)$ exists at $0$ is irrelevant. $\lim_{x \to a} \frac {f'(x)}{g'(x)}$ can exist when $\lim_{x \to a} f'(x)$ does not. Nov 30, 2020 at 20:36
• You're forgetting another condtion: $g(x)$ has to be nonzero in some small neighbourhood of $a$, except at $a$ itself. Nov 30, 2020 at 20:37

Hypothesis 1 only says that the functions need to be differentiable near $$a$$, not differentiable at $$a$$. Using symbols, they need only be differentiable on some deleted neighbourhood $$(a - \delta, a) \cup (a, a + \delta)$$ for some $$\delta > 0$$.
This is similar to what we think of for limits: $$\lim_{x \to a} F(x)$$ only depends on the behaviour of $$F(x)$$ as $$x$$ approaches $$a$$, and doesn't depend on $$F(a)$$ itself.