Question

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

Answer 1

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