Question

Математический анализ.Предел функции

Answer 1

$\lim_{x \to 1} \frac{sin(\pi x)(x^2+1)}{x^2-1}=[0/0]=$

$=\lim_{x \to 1} \frac{(sin(\pi x)(x^2+1))'}{(x^2-1)'}=\lim_{x \to 1} \frac{-(x^2+1)\pi cos(\pi x)+2xsin(\pi x)}{2x}=$

$= \lim_{x \to 1} \frac{-(1+1)\pi+0}{2}=-\pi$