Question

$\lim_{n \to \-1} \frac{2x^{2}+x-1 }{5x^{2}+4x-1 }$ ) Вычислите предел функции

Answer 1

0,5

Пошаговое объяснение:

$\lim_{x \to -1} \frac{2x^2+x-1}{5x^2+4x-1}=?\\\\2x^2+x-1=2(x-0,5)(x+1)\\D=9\\x_1=0,5\\x_2=-1\\\\5x^2+4x-1=5(x-0,2)(x+1)\\D=36\\x_1=0,2\\x_2=-1\\\\\\\lim_{x \to -1} \frac{2x^2+x-1}{5x^2+4x-1}=\lim_{x \to -1} \frac{2(x-0,5)(x+1)}{5(x-0,2)(x+1)}=\lim_{x \to -1} \frac{2(x-0,5)}{5(x-0,2)}=\\\\=\frac{2}{5} \lim_{x \to -1} \frac{x-0,5}{x-0,2}=\frac{2}{5}*\frac{-1-0,5}{-1-0,2}=\frac{2*(-1,5)}{5*(-1,2)}=\frac{-3}{-6}=0,5$