# Find the limit of $$\displaystyle{\left({e}^{{{2}{x}}}+{1}\right)}^{{\frac{{1}}{{x}}}}$$

Find the limit of ${\left({e}^{2x}+1\right)}^{\frac{1}{x}}$
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shanna87mn
We have ${e}^{2x}<{e}^{2x}+1<{e}^{2x}+{e}^{2x}$ for $x>0$. Hence, we get
${\left({e}^{2x}\right)}^{\frac{1}{x}}<{\left({e}^{2x}+1\right)}^{\frac{1}{x}}<{\left({e}^{2x}+{e}^{2x}\right)}^{\frac{1}{x}}$
i.e.
${e}^{2}<{\left({e}^{2x}+1\right)}^{\frac{1}{x}}<{\left(2{e}^{2x}\right)}^{\frac{1}{x}}={2}^{\frac{1}{x}}{e}^{2}$
Now take the limit and note that $\underset{x\to \mathrm{\infty }}{lim}{2}^{\frac{1}{x}}=1$ to get that
${e}^{2}\le \underset{x\to \mathrm{\infty }}{lim}{\left({e}^{2x}+1\right)}^{\frac{1}{x}}\le {e}^{2}$
Hence, we get that
$\underset{x\to \mathrm{\infty }}{lim}{\left({e}^{2x}+1\right)}^{\frac{1}{x}}={e}^{2}$