# Find the limit of the sequence or determine that the sequence diverges. <mroot> e

Find the limit of the sequence or determine that the sequence diverges.
$\sqrt[n]{{e}^{3n+4}}$
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Savion Stanton
Consider:
$\sqrt[n]{{e}^{3n+4}}$
Let $y=\left({e}^{3n+4}{\right)}^{\frac{1}{n}}$
Simplify this expression

So, $\mathrm{ln}\left(y\right)=\mathrm{ln}\left(\left({e}^{3n+4}{\right)}^{\frac{1}{n}}\right)$
Apply the property of the logarithm function $\mathrm{ln}\left({m}^{n}\right)=n\mathrm{ln}m$
We get,
$\mathrm{ln}\left(y\right)=\frac{1}{n}\mathrm{ln}\left({e}^{3n+4}\right)$
Apply the limit
$\underset{n\to \mathrm{\infty }}{lim}\mathrm{ln}\left(y\right)=\underset{n\to \mathrm{\infty }}{lim}\frac{1}{n}\mathrm{ln}\left({e}^{3n+4}\right)$
$\mathrm{ln}\left(y\right)=\underset{n\to \mathrm{\infty }}{lim}\left(\frac{1}{n}\mathrm{ln}\left({e}^{3n+4}\right)\right)$

After substituting limit we get the form of $\frac{\mathrm{\infty }}{\mathrm{\infty }}$

Apply the L’hospital Rule
$\mathrm{ln}\left(y\right)=\underset{n\to \mathrm{\infty }}{lim}\left(\frac{1\left(3\right)}{{e}^{3n+4}\left(1\right)}\right)$
$=\underset{n\to \mathrm{\infty }}{lim}\left(\frac{3}{{e}^{3n+4}}\right)$

Apply the limit
$\mathrm{ln}\left(y\right)=\frac{3}{{e}^{3\left(\mathrm{\infty }\right)+4}}$
$=\frac{3}{{e}^{\mathrm{\infty }}}$
$=0$

Apply the rule of logarithm function $\mathrm{ln}m=n⇒m={e}^{n}$
Here $m=y$ and $n=0$
$y={e}^{0}$
$y=1$
Thus, the limit is $1$