Is my justification right to evaluate the limit? I want to evaluate :
$$\lim_{x\to 0^+}\left(\frac1e(1+x)^{\tfrac1x}\right)^{\tfrac{1}{x^2}}$$
We know $\lim_{x\to 0^+}(1+x)^{\tfrac1x}$  is other form of $\lim_{u\to\infty}(1+\frac1u)^u=e$ which is increasing on $(0,\infty)$ it means at the infinity we get $e^-$ (the value is a litter less than $e$) therefor in the original limit we have $(1^-)^{\infty}$ therefore the limit equals to $0$.
Is my justification right?
 A: No, it is not. The limit $\lim_{u\to\infty}\left(1+\frac1u\right)^u$ is not “a litter less than $e$”; it is equal to $e$. And therefore $\lim_{x\to0^+}\frac1e(1+x)^{1/x}$ is exactly $1$. So, your limit is an indeterminate form and you will have to work harder to find its value.
Note: Your limit is indeed $0$, but, if your argument was correct, it would also prove that$$\lim_{x\to0^+}\left(\frac1e(1+x)^{1/x}\right)^{1/x}=0.$$However, this is not true; actually,$$\lim_{x\to0^+}\left(\frac1e(1+x)^{1/x}\right)^{1/x}=\frac1{\sqrt e}.$$
A: Another good way to think is :
Using Logarithm Identity - $ \ \ \displaystyle e^{\ln a} = a. \ \ $Thus,
$$\displaystyle  \left ( \frac {1}{e} (1+x)^{1/x}  \right )^{1/x^2} = \ \ e^{\displaystyle \frac {\ln (1+x)-x }{x^3}} .$$
Using Taylor Expansion, $ \ \ \displaystyle \ln (1+x) = x - \frac {x^2}{2} + \frac {x^3}{3} - \frac {x^4}{4} + ... $
Thus expression gets simplified to $ \ \ \displaystyle e^{ 1/3 - 1/2x }, $ as rest of the terms equals zero according to the limit.
$$\displaystyle \lim _{x \rightarrow 0} e^{ 1/3 - 1/2x } = 0.$$
