[Math] $\lim_{x\rightarrow 0^+} \frac {\ln(x)}{\ln( \sin x)}$ without l’Hôpital’s rule

Question

How to calculate $\displaystyle
\lim_{x\rightarrow 0^{+}}\frac{\ln x}{\ln (\sin x)}$ without l'Hôpital's rule please?
If anybody knows please help
I don´t have any idea 🙁
I´m looking forward your helps

Answer 1

If you are allowed to use the limit $\lim_{x\to0}{\sin x\over x}=1$, then you can say

$$\begin{align} \lim_{x\to0^+}{\ln(\sin x)\over\ln x}&=\lim_{x\to0^+}{\ln x+\ln({\sin x\over x})\over\ln x}\\ &=1+\left(\lim_{x\to0^+}\ln\left({\sin x\over x}\right)\right)\left(\lim_{x\to0^+}\left({1\over\ln x}\right)\right)\\ &=1+0\cdot0\\ &=1 \end{align}$$

(Note, I inverted the limit for the sake of simplicity. If the answer hadn't turned out to be $1$, it would be necessary to take its inverse.)

Added later: It occurs to me you really don't need to know the limit for $(\sin x)/x$. All you need is a pair of inequalities, such as

$${x\over2}\le\sin x\le x$$

(for small positive $x$) since that becomes

$$\ln x-\ln2\le\ln(\sin x)\le \ln x$$

so that

$$1\le{\ln(\sin x)\over\ln x}\le1-{\ln 2\over\ln x}$$

when $x\lt1$ (the inequalities reverse because you're dividing by a negative number), and the Squeeze Theorem now does the rest. The inequalities bounding $\sin x$ above and below are fairly easy to prove from the geometric definition of the sine function, interpreting $x$ as the arc length along the unit circle.