Question

Найди предел,используя метод Лопиталя.

Answer 1

$f(x)=(x+2^{x})^{frac{1}{x}}; ; ,; ; lnf(x)=frac{1}{x}cdot ln(x+2^{x})\\limlimits _{x to +infty}lnf(x)=limlimits _{x to +infty}frac{ln(x+2^{x})}{x}=Big [frac{infty }{infty }Big ]=limlimits _{x to +infty}frac{frac{1+2^{x}ln2}{x+2^{x}}}{1}=\\=limlimits _{x to +infty}frac{1+2^{x}ln2}{x+2^{x}}=Big [frac{infty }{infty }Big ]=limlimits _{x to +infty}frac{0+2^{x}ln^22}{1+2^{x}ln2}=Big [frac{infty }{infty }Big ]=\\=limlimits _{x to +infty}frac{2^{x}ln^32}{2^{x}ln^22}=limlimits _{x to infty}ln2=ln2$

$limlimits _{x to +infty}ln(x+2^{x})^{frac{1}{x}}=ln2; ; Rightarrow ; ; ; limlimits _{x to +infty}(x+2^{x})^{frac{1}{x}}=e^{ln2}=2$