# How do I prove that: <munder> <mo form="prefix">lim <mrow class="MJX-TeXAtom-ORD">

How do I prove that:
$\underset{n\to \mathrm{\infty }}{lim}\sqrt[n]{{4}^{n}+9{n}^{2}}=4$
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Raven Higgins
Squeeze theorem:
For $n\ge 4$
and so
$4\le \sqrt[n]{{4}^{n}+9{n}^{2}}\le {2}^{1/n}×4$
(We used $9{n}^{2}<{4}^{n}$ for $n\ge 4$, which has an easy proof using induction).
Since ${2}^{1/n}\to 1$ as $n\to \mathrm{\infty }$, the result follows.
To prove that ${2}^{1/n}\to 1$ one way to see this is to use the following standard theorem:
If ${a}_{n}>0$ and $\underset{n\to \mathrm{\infty }}{lim}\frac{{a}_{n+1}}{{a}_{n}}=L$, then $\underset{n\to \mathrm{\infty }}{lim}{a}_{n}^{1/n}=L$. You pick ${a}_{n}=2$. Of course, you could use this theorem on your original sequence itself...