$\lim_{x\rightarrow 0^+} \frac {\ln(x)}{\ln( \sin x)}$ 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.

Answer 2

Here is my approach but feel free to comment. Let $\frac{\ln x}{\ln(\sin x)}=t$ Then cross multiply gives $\ln x=t \ln(\sin x)=\ln(\sin x)^t$ This can be written as $x=(\sin x)^t$ after taking e-powers. So it follows $\frac{(\sin x)^t}{x}=1$ Recognizing a standard limit, the value for $t$ that works is if $t=1$ and so that would be the answer to the limit.

Answer 3

If you are allowed to use the Maclaurin series expansions (see this Wikipedia entry) for $\sin x$ and $\ln (1-x)$, then you can compute as follows. Since $$\sin x= x-\frac{1}{6}x^3+\frac{1}{120}x^5+O(x^7)$$

and

$$\ln(1-x)= -x-\frac{1}{2}x^2-O(x^3),$$

combining both, we have that

\begin{eqnarray*} L &=&\lim_{x\rightarrow 0^{+}}\frac{\ln x}{\ln (\sin x)}=\lim_{x\rightarrow 0^{+}}\frac{\ln x}{\ln x-\frac{1}{6}x^{2}-\frac{1}{180}x^{4}+O\left( x^{6}\right) } \\ &=&\left[ \lim_{x\rightarrow 0^{+}}\left( 1-\frac{x^{2}}{6\ln x}-\frac{x^{4}}{ 180\ln x}+\frac{1}{\ln x} O(x^{6})\right) \right] ^{-1} \\ &=&\left( 1+0+0+0\right) ^{-1} \\ &=&1. \end{eqnarray*}

Answer 4

$\frac{\ln x}{\ln \sin x }=\frac{\ln x-\ln \sin x}{\ln \sin x }+1=\frac{\ln \frac{x}{\sin x}}{\ln \sin x }+1\rightarrow \frac{\ln 1}{-\infty}+1=1$