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

Question

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?

See https://math.stackexchange.com/questions/75130/how-to-prove-that-lim-limits-x-to0-frac-sin-xx-1?noredirect=1&lq=1 — Robert Z, Mar 26 '19 at 10:23

score 6 · Accepted Answer · answered Mar 26 '19 at 10:24

6

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

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

answered Mar 26 '19 at 10:24

Fred

score 2 · Answer 2 · answered Mar 26 '19 at 10:26

2

Hint:

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

answered Mar 26 '19 at 10:26

5xum

2 Answers2