1
$\begingroup$

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}

  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, \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?

$\endgroup$
3
  • $\begingroup$ 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. $\endgroup$ Nov 30, 2020 at 20:35
  • $\begingroup$ 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. $\endgroup$
    – player3236
    Nov 30, 2020 at 20:36
  • $\begingroup$ You're forgetting another condtion: $g(x)$ has to be nonzero in some small neighbourhood of $a$, except at $a$ itself. $\endgroup$
    – Bernard
    Nov 30, 2020 at 20:37

1 Answer 1

1
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ How would I show that ln(x) and sin(x) are differentiable to the right hand side of zero? $\endgroup$ Nov 30, 2020 at 21:09

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .