Question

Решить задачи:

Answer 1

$\lim_{x \to -1} (4x^2+3)=4+3=7$
$\lim_{x \to -3} \frac{x^2-9}{x+3} =( \frac{0}{0} )= \lim_{x \to -3} \frac{(x-3)(x+3)}{x+3} =$$= \lim_{x \to -3}(x-3)=-3-3=-6$
$\lim_{x \to \infty} (1+ \frac{1}{x} )^{5x}=\lim_{x \to \infty} ((1+ \frac{1}{x} )^{x})^5=e^5$