Need a hint evaluating $ \lim\limits_{x\to 0}\frac{x\ln{(\frac{\sin (x)}{x})}}{\sin (x) - x} $

Question

I'm stuck with this. I've tried substituting $t$ for $\frac{\sin (x)}{x}$ and $\sin (x) - x$ but it doesn't work at all.

A small hint would be greatly appreciated.

Answer 1

$$\lim_{x\to 0}\frac{x\ln{(\frac{\sin (x)}{x})}}{\sin (x) - x} = \lim_{x\to 0}\frac{\ln{(1+\frac{\sin (x)-x}{x})}}{\frac{\sin (x) - x}{x}} =1 $$ Because $$\lim_{x\to0}\frac{\sin x-x}{x}=\lim_{x\to0}\frac{\sin x}{x}-1=1-1=0$$ and $$\lim_{t\to0}\frac{\ln(1+t)}{t}=1.$$

Answer 2

Use the Taylor expansion $$ \frac{\sin x}{x}=1-\frac{x^2}{6}+o(x^2) $$ to conclude that $$ x\ln\frac{\sin x}{x}=-\frac{x^3}{6}+o(x^3) $$ Likewise $$ \sin x-x=x-\frac{x^3}{6}+o(x^3)-x=-\frac{x^3}{6}+o(x^3) $$ So the limit is $1$.