Question

Помогите решить предел!

Answer 1

$\displaystyle \lim_{x \to \infty} (\frac{x+2}{x})^{3-2x}=1^\infty=[\lim_{x \to \infty}(1+\frac{1}{\frac{x}{2}})^\frac{x}{2}]^{\displaystyle\lim_{x \to \infty} {\frac{6-4x}{x}}}=e^{-4}=\frac{1}{e^4}$