# L'Hôpital's rule - How solve this limit question

How to solve this ? $$\lim_{x\to 0} f(x)\;\text{where}\;f(x)=\frac{ \arctan(2x)}{\ln (x)}$$

The answer is $$0$$. My question is when we plug in $$0$$ in $$f(x)$$, we get the form $$\frac{0}{\infty}$$, which is not an indeterminate form, so we might just write $$0$$ as answer directly OR if we apply L'Hôpital's rule, we would still get an answer as $$0$$. Which method is correct?

• We can't use L'Hôpital's rule for the reasons you mentioned, so that approach isn't correct. You are essentially right with the first explanation. To see this more clearly think of $f(x) = g(x)h(x)$, where $g(x) = arctan(2x)$, and $h(x) = 1/ln(x)$. Do you know how to do these two limits separately? – Jabbath Nov 14 '18 at 6:21
• First one because L-Hopital's rule is only applicable when you get an indeterminate form. You can get different answers if the limits are not indeterminate. – Krishna Nov 14 '18 at 6:22

First of all, the limit cannot be $$x\to 0$$, it must be $$x\to 0^+$$, because of the domain of ln(x).
Second thing is that the L-Hopital rule is not applicable in this case, because it applies only for $$\frac{0}{0}\ or\ \frac{\infty}{\infty}$$ form, so the first method is correct.
Note that we can only consider $$\lim_{x\to 0^+} f(x)$$ and that we have a $$\frac 0{-\infty}$$ form which is not indeterminate.