0
$\begingroup$

Evaluate $\lim\limits_{x\to 0^{+}}(\ln(x)-\ln(\sin x))$

My trial

As $x\to 0^{+},\;\ln(x)\to \infty$ and $\ln(\sin x)\to \infty.$ So,

\begin{align}\lim\limits_{x\to 0^{+}}(\ln(x)-\ln(\sin x))=\lim\limits_{x\to 0^{+}}\ln\left(\frac{x}{\sin x}\right)\end{align}

This should result to $\infty$ but I may be wrong. If I am wrong, how do I apply L'Hopital's rule to this?

$\endgroup$
5
$\begingroup$

$ \frac{x}{\sin x} \to 1$ as $x \to 0$,

hence $ \ln (\frac{x}{\sin x}) \to \ln 1=0$ as $x \to 0.$

$\endgroup$
  • $\begingroup$ Oh, thanks Fred! $\endgroup$ – Omojola Micheal Mar 26 at 10:25
2
$\begingroup$

Hint:

  1. If $f$ is a continuous function and $\lim_{x\to a} g(x) = L$, then $\lim_{x\to a} f(g(x)) = f(L)$.
  2. $\ln$ is continuous.
  3. $\lim_{x\to 0}\frac{x}{\sin x}$ is simple to calculate.
$\endgroup$

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.