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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?

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  • $\begingroup$ 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? $\endgroup$ – Jabbath Nov 14 '18 at 6:21
  • $\begingroup$ 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. $\endgroup$ – Krishna Nov 14 '18 at 6:22
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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.

Hope it is helpful.

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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.

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