I have that the following hypothesis need to be satisfied in order to use L'Hospital's rule for,\begin{equation} \lim_{x\to a}\frac{f(x)}{g(x)} \end{equation}
$f(x)$ and $g(x)$ have to be functions differentiable near $a$.
$\lim_{x\to a}\frac{f(x)}{g(x)}$ has to have the indeterminate form $\frac{0}{0}$ or $\frac{\pm\infty}{\pm\infty}$.
$\lim_{x\to a}\frac{f'(x)}{g'(x)}$ has to exist.
If possible, find the following limit using L’Hospital’s rule. If not possible, explain why,\begin{equation} \lim_{x\to0^+}\frac{\ln{x}}{\frac{1}{\sin{x}}} \end{equation}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?
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$\begingroup$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.
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