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

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?

• – Robert Z Mar 26 at 10:23
• @Robert Z: Thanks a lot, I got that hint! – Omojola Micheal Mar 26 at 10:26

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