<s>[INST] Finding the limit of $\left(\frac{(1+2x)^{1/x}}{e^2}\right)^{1/x}$ at $x=0$ I can't seem to find a solution to this.
$$\lim_{x\to0} \left(\frac{(1+2x)^{1/x}}{e^2}\right)^{1/x}$$
i tried to manipulate to apply Lhopitals rule but i can't see to do it
 [/INST] There have been many nice solutions provided by others earlier. 
This is an alternative approach, without using logs or l'Hopital:
$$\begin{align}
\dfrac {\left(1+\dfrac kn \right) ^n}{e^k}
&= {\left(1+\dfrac kn \right) ^n}e^{-k}\\
&={\left(1+\dfrac kn \right) ^n}\left(1-k+\frac{k^2}{2!}-\frac{k^3}{3!}+... \right)\\
&=\left[ 1+n\left(\frac kn\right)+\dfrac{n(n-1)}{2!}\left(\frac kn \right)^2+\cdots \right] \left[1-k+\frac{k^2}{2!}-\frac{k^3}{3!}+\cdots  \right]\\
&\approx \left[1+k+\frac{n-1}{2n}k^2 \right]\left[ 1-k+\frac{k^2}2\right]\\
&\approx 1+k-k+k^2 \left(\frac{n-1}{2n}-1+\frac 12 \right) \\
&\approx 1-\frac{k^2}{2n}\\
\lim\limits_{n\to\infty} \left(\dfrac{\left(1+ \dfrac kn\right)^n}{e^k}\right)^n
&=\lim\limits_{n\to\infty} \left(1-\dfrac{k^2}{2n}\right)^n\\
&=e^{-k^2/2}
\end{align}$$
Put $k=2$ and $n=\dfrac 1x$:
$$\begin{align}
\lim\limits_{x\to 0} 
\left(\dfrac{\left(1+ 2x \right)^{\frac 1x}}{e^2}\right)^{\frac 1x}
&=e^{-2}\end{align}$$
</s>