[Edit] The other answers are much better. But, as this falls into a standard class of problems in the application of L'Hopital's rule, I'll offer:
If you have an $\infty^0$ form, it is indeterminate (for example $\lim\limits_{x\rightarrow\infty}(e^x)^{1/x}=e$, but $\lim\limits_{x\rightarrow\infty}((e^x)^x)^{1\over x}=\infty$).
To handle a form of this type, you may take the limit of the logarithm of the expressionn first. Then, using the power rule for logarithms, you can set things up so that L'Hopital's rule is applicable. After you find the limit of the logarithm, say it's $L$, then the limit of the original expression will be $e^L$.
For your example:
Find the limit of the logarithm first.
Set $g(x)=\ln\bigl( (e^{2x}+1)^{1/x}\bigr)= \underbrace{{1\over x}\ln( e^{2x}+1)}_{0\cdot\infty\text{ form}}= \underbrace{{\ln (e^{2x}+1)\over x}}_{{\infty\over\infty}\text{ form}}.$
Then, using L'Hopital: $\lim_{x\rightarrow\infty} g(x)= \lim_{x\rightarrow\infty}{\ln(e^{2x}+1)\over x}=\lim_{x\rightarrow\infty}{2e^{2x} \over e^{2x}+1} =\lim_{x\rightarrow\infty}{4e^{2x} \over 2e^{2x} } =2. $
So $\eqalign{ \lim_{x\rightarrow\infty}{(e^{2x}+1 )^{1/x}} &=\lim_{x\rightarrow\infty}{ \exp\bigl[\, g(x) \bigr]}\cr &={ \exp\bigl[\, \lim_{x\rightarrow\infty} g(x) \bigr]}\cr & =e^2.} $