Evaluating the limit: $\lim \limits_{n\to \infty}\sqrt[n]{4^n+9n^2}=4$ How do I prove that:
 $\lim \limits_{n\to \infty}\sqrt[n]{4^n+9n^2}=4$
Thank you.
 A: Just for fun:

Use a sledgehammer:
You could apply L'Hôpital's rule to $\ln\root n\of{4^n+9n^2}$:
We have
$$\eqalign{
\lim_{n\rightarrow\infty}\ln\root n\of{4^n+9n^2} 
&=\lim_{n\rightarrow\infty}{\ln(4^n+9n^2)\over n}\cr
&=\lim_{n\rightarrow\infty}{ {\ln 4\cdot4^n+18n \over 4^n+9n^2 }}\cr
&=\lim_{n\rightarrow\infty}{ {(\ln 4)^2\cdot4^n+18  \over \ln 4\cdot4^n+18n }}\cr
&=\lim_{n\rightarrow\infty}{ {(\ln 4)^3\cdot4^n   \over (\ln 4)^2\cdot4^n+18  }}\cr
&=\lim_{n\rightarrow\infty}{ {(\ln 4)^4\cdot4^n   \over (\ln 4)^3\cdot4^n  }}\cr
&= { {\ln 4 }};\cr
}
$$
whence $\lim\limits_{n\rightarrow\infty} \root n\of{4^n+9n^2} =e^{\ln 4}=4.$

Or, use an even even bigger sledgehammer:
Use the fact that for positive $a_n$ if $\lim\limits_{n\rightarrow\infty} {a_{n+1}\over a_n}$ exists then so does $\lim\limits_{n\rightarrow\infty}\root n\of {a_n}$ and they are equal.
Here $a_n=4^n+9n^2$ and one can show
$$
\lim\limits_{n\rightarrow\infty} {4^{n+1}+9(n+1)^2\over 4^n+9n^2} =4.
$$
So, then  $\lim\limits_{n\rightarrow\infty} \root n\of{4^n+9n^2}=4$ as well.
A: We are looking at
$$4\left(1+\frac{9n^2}{4^n}\right)^{1/n}.$$
Then use the Squeeze Theorem.
Remark: This approach has the (small!) advantage that we need to know essentially nothing about $n$-th root, apart from the fact that the $n$-th root of $x$ is $\le x$ if $x\ge 1$.  All we need to know is that $\dfrac{9n^2}{4^n}$ can be made "small." 
A: Detailed hint: 


*

*Write $\sqrt[n]{4^{n}+9n^{2}}$ as $$\sqrt[n]{4^{n}+9n^{2}}=4\sqrt[n]{ 1+9n^{2}/4^{n}}.$$  Answering your comment above: Why? Because $$\sqrt[n]{4^{n}+9n^{2}}=\sqrt[n]{4^{n}\left( 1+9n^{2}/4^{n}\right) }=4\sqrt[n]{ 1+9n^{2}/4^{n}}.$$

*Observe that $$\lim_{n\rightarrow \infty }\frac{n^{2}}{4^{n}}=0.$$
See this question How to prove that exponential grows faster than polynomial?
A: Squeeze theorem:
For $n \ge 4$ we have that
$$ \sqrt[n]{4^n} \le \sqrt[n]{4^n + 9n^2} \le \sqrt[n]{2\times4^n}$$
and so
$$ 4 \le \sqrt[n]{4^n + 9n^2}  \le 2^{1/n} \times 4 $$
(We used $9n^2 \lt 4^n$ for $n \ge 4$, which has an easy proof using induction).
Since $2^{1/n} \to 1$ as $n \to \infty$, the result follows.
To prove that $2^{1/n} \to 1$ one way to see this is to use the following standard theorem:

 If $a_n \gt 0$ and $\lim_{n \to \infty} \frac{a_{n+1}}{a_n} = L$, then $\lim_{n \to \infty} a_n^{1/n} = L$. You pick $a_n = 2$. Of course, you could use this theorem on your original sequence itself...

