How to calculate $\lim_{x\to0}(x+e^{2x})^{\frac2 x}$

Question

$$\lim_{x\to0}(x+e^{2x})^{\frac2 x}$$ I saw somewhere that they put all the expression as the power of e with ln but I can't figure why.

Answer 1

$\lim_{x\to0}\frac{e^{2x}-1}x=2$, so

$$\lim_{x\to0}(x+e^{2x})^{\frac2 x}=\lim_{x\to0}(1+(x+e^{2x}-1))^{\frac1{x+e^{2x}-1}(2+\frac{2(e^{2x}-1)}x)}=e^6 $$

Answer 2

A related technique. Here is an approach $$ e^{2/x\ln(x+e^{2x})}= e^{2/x(3x-5/2x^2+\dots)} \longrightarrow_{x\to 0} e^6.$$

Note: We used the Taylor series

$$\ln(x+e^{2x})= 3\,x-{\frac {5}{2}}{x}^{2}+O \left( {x}^{3} \right) $$.