Question

multumesc anticipat ​

Answer 1

Răspuns:

$\lim_{x\to -\infty} e^x(x-3)^2 = \lim_{x\to -\infty} \frac{(x-3)^2}{e^{-x}} =$ (se aplica L'Hospital, cazul $\frac{\infty}{\infty}$) $= \lim_{x\to -\infty}\frac{2(x-3)}{-e^{-x}} = \lim_{x\to -\infty}\frac{2}{e^{-x}} = 0$.