Question

Помогите решить предел

Answer 1

$limlimits _{x to infty}frac{(2x+1)^{x+2}}{(2x-1)^{x+2}}=limlimits _{x to infty}Big (1+frac{2}{2x-1}Big )^{x+1}=limlimits_{x to infty}Big (1+frac{2}{2x-1}Big )^{frac{2x-1}{2}cdot frac{2cdot (x+1)}{2x-1}}=\\=limlimits _{x to infty}Big (underbrace {Big (1+frac{2}{2x-1}Big )^{frac{2x-1}{2}}}_{to e}Big )^{frac{2x+2}{2x-1}}=e^{limlimits _{x to infty}frac{2x+2}{2x-1}}=e^1=e$