I need to calculate the below limit:
$$ \lim_{x\to1^+} \frac{\ln(x-1)}{\cot(x-1)} $$
My steps are as follows:
$$ \lim_{x\to1^+} \frac{\frac{1}{x-1}}{-\frac{1}{\sin^2(x-1)}} = -\lim_{x\to 1^+} \frac{\sin^2(x-1)}{x-1}=-\lim_{x\to1^+}\frac{2\cos(x-1)\sin(x-1)}{1}=0 $$
Is my solution correct?