Question

Решите предел в пределе -5 не входит в корень
$\lim_{x \to \-5} \frac{\sqrt{5-4x}-5 }{x^{2}-25 }$

Answer 1

Ответ:

$\displaystyle \lim_{x \to -5}\dfrac{\sqrt{5-4x}-5}{x^2-25}=\lim_{x \to -5}\dfrac{(\sqrt{5-4x}-5)(\sqrt{5-4x}+5)}{(x-5)(x+5)(\sqrt{5-4x}+5)}=\\\\\\=\lim_{x \to -5}\dfrac{(5-4x)-25}{(x-5)(x+5)(\sqrt{5-4x}+5))}=\lim_{x \to -5}\dfrac{-4(x+5)}{(x-5)(x+5)(\sqrt{5-4x}+5))}=\\\\\\=\lim_{x \to -5}\dfrac{-4}{(x-5)(\sqrt{5-4x}+5)}=\dfrac{-4}{-10\cdot 10}=\dfrac{4}{100}=0,04$