Evaluating the limit: $\lim \limits_{n\to \infty}\sqrt[n]{4^n+9n^2}=4$

Question

How do I prove that: $\lim \limits_{n\to \infty}\sqrt[n]{4^n+9n^2}=4$

Thank you.

Answer 1

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...

Answer 2

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.

Answer 3

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."

Answer 4

Detailed hint:

1. 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 $$\sqrt[n]{4^{n}+9n^{2}}=\sqrt[n]{4^{n}\left( 1+9n^{2}/4^{n}\right) }=4\sqrt[n]{ 1+9n^{2}/4^{n}}.$$
2. Observe that $$\lim_{n\rightarrow \infty }\frac{n^{2}}{4^{n}}=0.$$ See this question How to prove that exponential grows faster than polynomial?