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?


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

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

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


  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.

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