Question

Помогите пожалуйста найти пределы

Answer 1

$\displaystyle\lim_{x\to0}(2-cos(x))^{\displaystyle\frac{1}{sin^2(x)}}=1^{\displaystyle\infty}=\lim_{x\to0}(1+1-cos(x))^{\displaystyle\frac{1}{sin^2(x)}}=\\=[\lim_{x\to0}(1+1-cos(x))^{\displaystyle\frac{1}{1-cos(x)}}]^{\displaystyle\frac{1-cos(x)}{sin^2(x)}}=e^{\displaystyle\lim_{x\to0}\frac{1-cos(x)}{sin^2(x)}}=\\=e^{\displaystyle\lim_{x\to0}\frac{x^2}{2*x^2}}=e^{\displaystyle\frac{1}{2}}=\sqrt e$

$\displaystyle\lim_{x\to0}(1+5x)^{\displaystyle\frac{1}{2x+3}}=1^{\displaystyle\frac{1}{3}}=1$

$\displaystyle\lim_{x\to3+0}4^{\displaystyle\frac{1}{3-x}}=4^{\displaystyle-\infty}=0$