solving the limit of $(e^{2x}+x)^{1/x}$ I tried to solve for the following limit: 
$$\lim_{x\rightarrow \infty} (e^{2x}+x)^{1/x}$$
and I reached to the indeterminate form: 
$${4e^{2x}}\over {4e^{2x}}$$ 
if I plug in, I will get another indeterminate form! 
 A: Tricky solution:
$$\begin{align}
(e^{2x}+x)^{1/x}&=e^{2x/x}\left(1+\frac{1}{e^{2x}/x}\right)^{1/x}\\
&=e^2\left(\left(1+\frac{1}{e^{2x}/x}\right)^{e^{2x}/x}\right)^{e^{-2x}}
\end{align}$$
Since $e^{2x}/x\to\infty$ and $(1+1/y)^y\to e$ as $y\to\infty$, and $e^{-2x}\to 0$, you get the limit is $e^2$.
A: I have no idea how you got $\lim\limits_{x\to\infty}\frac{4e^{2x}}{4e^{2x}}$, but as has been pointed out, that limit is easily evaluated: the fraction is identically $1$, so the limit is also $1$.
Let $L=\lim\limits_{x\to\infty}\left(e^{2x}+x\right)^{1/x}$; then 
$$\ln L=\ln\lim_{x\to\infty}\left(e^{2x}+x\right)^{1/x}=\lim_{x\to\infty}\ln\left(e^{2x}+x\right)^{1/x}=\lim_{x\to\infty}\frac{\ln\left(e^{2x}+x\right)}x\;.$$
Now apply l’Hospital’s rule.
A: Another way, considering $$A=(e^{2x}+x)^{1/x}$$ Taking logarithms $$\log(A)=\frac 1x \log(e^{2x}+x)=\frac 1x \left(\log(e^{2x})+\log(1+\frac x {e^{2x}})\right)=\frac 1x \left(2x+\log(1+\frac x {e^{2x}})\right)$$ $$\log(A)=2+\frac 1x \log(1+\frac x {e^{2x}})\approx 2+\frac 1x \frac x {e^{2x}}=2+\frac 1 {e^{2x}}$$
A: $x>0\implies e^x>x \implies e^2=(e^{2 x})^{1/x}<(e^{2x}+x)^{1/x}<    (2 e^{2x})^{1/x}    =(e^2) (2^{1/x})$..... And $\lim_{x\to \infty}2^{1/x}=1.$
A: We simply have
$$(e^{2x}+x)^{1/x}=e^{2x/x}\left(1+\frac{x}{e^{2x}}\right)^{1/x}\to e^2\cdot1^0=e^2$$
