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! 

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.
