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Use L'Hôpital's rule to solve

$$\lim_{x\to 0^{+}}\sin(x)\ln(x)$$

My attempt:

$$\lim_{x\to 0^{+}}\sin(x)\ln(x) = \lim_{x\to 0^{+}} \frac{\ln(x)}{\sin(x)}$$

$\frac{\ln(0)}{\sin(0)}$ is in the form $\frac{-\infty}{0}$, it is indeterminate, and as such, using L'Hôpital's rule:

$$\lim_{x\to 0^{+}} \frac{\ln(x)}{\sin(x)} = \lim_{x\to0^{+}}\frac{1}{x\cos(x)}$$

I would have applied L'Hôpital's rule again, but to my horror, I realise that $\frac{1}{0}$ is not an indeterminate form, according to Wikipedia.

I realized that my reasoning, while it could let me get the correct answer, is wrong! How do you solve this question now?

EDIT: Is $\frac{-\infty}{0}$ indeterminate as well? I couldn't find it in Wikipedia. If it is not indeterminate, I couldn't use L'Hôpital's rule too!

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    $\begingroup$ This is simply a product of two standard limits $\lim_{x\to 0}(\sin x) /x=1$ and $\lim_{x\to 0^{+}}x\log x =0$ so that the answer is $0$. $\endgroup$ – Paramanand Singh Apr 30 '17 at 8:44
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It would be re-written as : $$\lim_{x \to 0^+} \frac{\ln (x)}{\csc (x)}$$

Instead of $$ \frac{\ln(x)}{\sin(x)}$$ Now use L'Hopital's rule.

Also note that the form $\dfrac{\infty}{0}$ isn't indeterminate, it already tends to $\infty$. The problem in your solution is that you accidentally wrote : $$\frac{1}{\sin x}= \sin x$$

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  • $\begingroup$ Hi, thanks. I guess I was careless. But my main concern now is whether you could apply L Hopital Rule for forms $\frac{-\infty}{0}$ and $\frac{1}{0}$. Can you do so? $\endgroup$ – Kyoma Apr 30 '17 at 8:44
  • $\begingroup$ @Kyoma No, L'HOPITAL'S rule applys only to indeterminate forms such as $\frac{0}{0}$ and $\frac{\infty}{\infty}$ $\endgroup$ – Jaideep Khare Apr 30 '17 at 8:45
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$$\lim_{x\to0^{+}} \frac{\ln(x)}{\csc(x)} = \lim_{x\to0^{+}} \frac{-1}{x\csc(x)\cot(x)} = \lim_{x\to0^{+}} \frac{-\sin(x)}{x} \lim_{x\to0^{+}} \tan(x) = 0$$

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