Question: 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

Answer: 
Whenever you see an exponent, use logs.
$$\require{cancel}\begin{align}y&=\left(\frac{(1+2x)^{\frac1{x}}}{e^2}\right)^\frac1{x} \\ \ln y &=\frac1{x} \ln\left(\frac{(1+2x)^{\frac1{x}}}{e^2}\right)=\frac1{x} \left(\underbrace{\ln(1+2x)^{\frac1{x}}}\_{\frac1{x}\ln(1+2x)}- \cancelto{2}{\ln (e^2)}\right)\\&=\frac1{x} \left(\frac1{x}\ln(1+2x)- 2\right)\\&=\frac{\ln(1+2x)-2x}{x^2}\end{align}$$

Use L'Hôspital, twice:

$$\begin{cases}\text{first: }&\frac{\frac2{1+2x}-2}{2x}=\frac{-4x}{2x+4x^2}\\\text{second:}&\frac{-4}{2+8x}\end{cases}$$

Now it's solveable, plug in $0$ for $x$ and you get $\frac{-4}{2}=-2$ Recall that this is equal to $\ln y$, hence:

$$ \ln y=-2 \\ \therefore y = e^{-2}$$
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 $n=\dfrac 1x$:

$$\begin{align}
\lim\limits\_{x\to 0} 
\left(\dfrac{\left(1+ 2x \right)^{\frac 1x}}{e^2}\right)^{\frac 1x}
&=e^{-2}\end{align}$$