How can I understand these limits?

Question

I would like to see a proof of these limits:

$$ \lim_{x\to 1^-} \left(\sin x\right)^{\frac{1}{1-x}}=0,\qquad \lim_{x\to 1^+} \left(\sin x\right)^{\frac{1}{1-x}}=+\infty. $$

Any help is appreciated.

Answer 1

$\log\lim_{x\to 1^-} \left(\sin x\right)^{\frac{1}{1-x}}=\lim_{x\to 1^-} \log\left(\sin x\right)^{\frac{1}{1-x}}=\lim_{x\to 1^-}\left(\dfrac{1}{1-x}\right)\log\sin x=\infty\cdot \log\sin 1=-\infty$

If the $\log$ tends to $-\infty$ the limit is $0$ because $e^x\to 0$ as $x\to \infty$

In a similar way you can do for the other limit. Be careful because $\lim_{x\to 1^+}\dfrac{1}{1-x}=-\infty$