Question

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$\lim_{x \to 1} \frac{sin(\pi*x^a)}{sin(\pi*x^b)}$

Answer 1

$\displaystyle \lim_{x \to 1} \frac{\sin(\pi x^a)}{\sin(\pi x^b)}=\lim_{x \to 1}\frac{\sin\pi(x^a-1+1)}{\sin \pi(x^b-1+1)}=\lim_{x \to 1}\frac{\sin(\pi+\pi(x^a-1))}{\sin(\pi+\pi(x^b-1))}=\\ \\ \\ =\lim_{x \to 1}\frac{\sin \pi(x^a-1)}{\sin\pi(x^b-1)}=\lim_{x \to 1}\frac{\pi(x^a-1)}{\pi(x^b-1)}=\left\{\begin{array}{ccc}t=x-1\\ x=t+1\\ t\to 0\end{array}\right\}=\\ \\ \\ =\lim_{t \to 0}\frac{(1+t)^a-1}{(1+t)^b-1}=\lim_{t \to 0}\frac{at+o(t)}{bt+o(t)}=\frac{a}{b}$