Question

Прошу помощи! Найдите пределы

Answer 1

Ответ:

$\lim_{n \to 3} \frac{9-x^2}{\sqrt{3x-3}} = \lim_{x \to 3} \frac{9-9}{\sqrt{9-3}} = 0$

$\lim_{x \to \pi+0} \frac{\sqrt{1 + cos(x)}}{sin(x)} = \lim_{x \to \pi+0} \frac{\sqrt{1 + cos(x)}*\sqrt{1-cos(x)}}{sin(x)*\sqrt{1-cos(x)}} = \lim_{x \to \pi+0} \frac{\sqrt{1-cos^2(x)}}{sin(x)*\sqrt{1-cos(x)}} = \lim_{x \to \pi+0} \frac{\sqrt{sin^2(x)}}{sin(x)*\sqrt{1-cos(x)}} = \lim_{x \to \pi+0} \frac{1}{\sqrt{1-cos(x)}} = \lim_{x \to \pi+0} \frac{1}{\sqrt{1-cos(\pi + 0)}} = \lim_{x \to \pi+0} \frac{1}{\sqrt{1+1}} = \frac{1}{\sqrt{2}}$

Объяснение: