Question

Найти предел функции:

Answer 1

$lim_{x to -1} frac{2 x^{2} +5x+3}{x+ x^{2} } = frac{2*(-1) ^{2}+5*(-1)+3 }{-1+(-1) ^{2} }= frac{0}{0}$
2x²+5x+3=0. D=1. x₁=-3/2. x₂=-1
2x²+5x+3=2*(x-(-3/2))*(x-(-1))=2*(x+3/2)*(x+1)=(2x+3)*(x+1)
$lim_{x to -1} frac{2 x^{2} +5x+3}{x+ x^{2} } = lim_{ to -1} frac{(2x+3)*(x+1)}{x*(x+1)} = lim_{x to -1} frac{2x+3}{x} =$
$= frac{2*(-1)+3}{-1} = frac{1}{-1}=-1$