Question

Высшая математика!!!
Вычисление пределов с бесконечностью

Answer 1

$\lim\limits _{x \to \infty}\Big (\frac{x+8}{x-7} \Big )^{4x-7}=[\; 1^{\infty }\, ]=\lim\limits _{x \to \infty}\Big (\underbrace {\Big (1+\frac{15}{x-7}\Big )^{\frac{x-7}{15}}}_{\to e}\Big )^{\frac{15}{x-7}\cdot (4x-7)}=\\\\=e^{\lim\limits _{x \to \infty}\frac{60x-105}{x-7}}=e^{60}\\\\\\\star \; \; \frac{x+8}{x-7}=\frac{(x-7)+7+8}{x-7}=\frac{x-7}{x-7}+\frac{7+8}{x-7}=1+\frac{15}{x-7}\; \; \star \\\\\star \; \; \lim\limits _{\alpha (x) \to \infty}\Big (1+\frac{1}{\alpha (x)}\Big )^{\alpha (x)}=e\; \; \star$