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
 A: 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:
$$ \ln y=-2 \\ \therefore y = e^{-2}$$
A: Once you wrote, as Shavar answered, $$\log(y)=\frac{\log(1+2x)-2x}{x^2}$$ you can also use Taylor series around $x=0$. This gives $$\log(1+2x)=2 x-2 x^2+\frac{8 x^3}{3}+O\left(x^4\right)$$ and then $$\log(y)=-2+\frac{8 x}{3}+O\left(x^2\right)$$ which again approximate $y$ as $$y=\frac{1}{e^2}+\frac{8 x}{3 e^2}+O\left(x^2\right)$$
A: $$\lim_{x\to0} \left(\frac{(1+2x)^{1/x}}{e^2}\right)^{1/x}\\
=\lim_{x\to0} \exp\left[(1/x)\ln\left(\frac{(1+2x)^{1/x}}{e^2}\right)\right]\\
=\lim_{x\to0} \exp\left[\frac1x\{(1/x)\ln(1+2x)-\ln(e^2)\}\right]\\
I=\lim_{x\to0} \exp\left\{\frac{\ln(1+2x)}{x^2}-\frac2{x}\right\}\\
I=\lim_{x\to0} \exp\left(4\left\{\frac{\ln(1+x)-x}{x^2}\right\}\right)$$
Let $$J=\lim_{x\to0}\left\{\frac{\ln(1+x)-x}{x^2}\right\}=\lim_{x\to0}\frac{\ln(1+x)}{x^2}-\frac1x\\
J=\lim_{x\to0}\frac{\ln(1+2x)}{4x^2}-\frac1{2x}\implies 2J=\lim_{x\to0}\frac{\ln(1+2x)}{2x^2}-\frac1{x}\\
J=\lim_{x\to0}\left(\frac{\ln(1+2x)}{2x^2}-\frac1{x}\right)-\left(\frac{\ln(1+x)}{x^2}-\frac1x\right)\\
J=\lim_{x\to0}\frac{\ln\frac{(1+2x)}{(1+x)^2}}{2x^2}=\lim_{x\to0}\frac1{2x^2}\ln\left(1-\frac{x^2}{1+x^2}\right)\\
J=\lim_{x\to0}\frac1{2x^2}\ln\left(1-\frac{x^2}{1+x^2}\right).\frac{-(x^2)/(1+x)^2}{-(x^2)/(1+x)^2}\\
=\lim_{x\to0}\frac{-(x^2)/(1+x)^2}{2x^2}=-\frac12
$$
Since $\lim_{x\to0}\ln(1+x)/x=1$
So, $$I=e^{4\times-\frac12}=e^{-2}$$
A: 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 $n=\dfrac 1x$:
$$\begin{align}
\lim\limits_{x\to 0} 
\left(\dfrac{\left(1+ 2x \right)^{\frac 1x}}{e^2}\right)^{\frac 1x}
&=e^{-2}\end{align}$$
