Question

решение пределов по математике

Answer 1

$1)\; \; \lim\limits _{x \to 0}\Big (\frac{1}{x+1}\Big )^{\frac{1}{2x}}=[\, 1^{\infty }\, ]= \lim\limits _{x \to 0}\Big (1+\frac{1}{x+1}-1\Big )^{\frac{1}{2x}}=\\\\= \lim\limits _{x \to 0}\Big (1+\frac{-x}{x+1}\Big )^{\frac{1}{2x}}= \lim\limits _{x \to 0}\Big (\Big (1+\frac{-x}{x+1}\Big )^{\frac{x+1}{-x}}\Big )^{\frac{-x}{2x(x+1)}}=\\\\=e^{ \lim\limits _{x \to 0}\, \frac{-x}{2x(x+1)}}=e^{\lim\limits _{x \to 0}\, \frac{-1}{2(x+1)}}=e^{-\frac{1}{2}}= \frac{1}{\sqrt{e}}$

$2)\; \; \lim\limits _{x \to 0}\frac{cos5x-cos7x}{x^2}= \lim\limits _{x \to 0}\frac{2sin6x\cdot sinx}{x^2}=\\\\=2\cdot \lim\limits _{x \to 0}\Big (6\cdot \underbrace {\frac{sin6x}{6x}}_{1}\cdot \underbrace{\frac{sinx}{x}}_{1}\Big )=2\cdot 6=12$