Question

Математический анализ.Предел функции

Answer 1

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

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

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