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!

  • 1
    $\begingroup$ 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

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 : $$\frac{1}{\sin x}= \sin x$$

  • $\begingroup$ 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

$$\lim_{x\to0^{+}} \frac{\ln(x)}{\csc(x)} = \lim_{x\to0^{+}} \frac{-1}{x\csc(x)\cot(x)} = \lim_{x\to0^{+}} \frac{-\sin(x)}{x} \lim_{x\to0^{+}} \tan(x) = 0$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.