How to solve this limit: $\lim\limits_{x \to 0}\left(\frac{(1+2x)^\frac1x}{e^2 +x}\right)^\frac1x$ $$\lim\limits_{x \to 0}\left(\frac{(1+2x)^\frac{1}{x}}{e^2 +x}\right)^\frac{1}{x}=~?$$
Can not solve this limit, already tried with logarithm but this is where i run out of ideas. Thanks.

Let us first examine the inner part.
$$\lim_{x \to 0} \frac{(1+2x)^{\frac{1}{x}}}{e^2+x}$$
It is easy to see that numerator is of the form $1^{ \infty}$ So the numerator goes to $e^2$.
Rewritting the limit $$\lim_{x \to 0}\frac{1}{e^{2/x}}(\frac{e^2}{e^2+x})^{\frac{1}{x}} (1+2x)^{1/x^2}$$ which can be rewritten as $$e^{-e^{-2}} \lim_{x \to 0}[\frac{(1+2x)^{1/x}}{e^2}]^{1/x}$$
Now evaluating the limit under the inner part.take ln both sides and use Lhospital rule we get the limit as $$e^{-2-e^{-2}}$$.In other way notice that it is in the form $1^{\infty}$.So the limit can be written as $$e^{\lim_{x \to 0} \frac{(1+2x)^{1/x}-e^2}{e^2}(1/x)}$$.Now application of lhospitals rule becomes easier.
