# 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.

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For the $\sqrt[n]{}$ you may want to use $\sqrt[n]{}$. – user21436 Mar 28 '12 at 17:22
yep, changed it, thanks. – Anonymous Mar 28 '12 at 17:23

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...

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$\lim \limits_{n\to \infty}\sqrt[n]{4^n}$=4 easy. However, why $\sqrt[n]{4^n + 9n^2} \le \sqrt[n]{2\times4^n}$ and why $\lim \limits_{n\to \infty}\sqrt[n]{2\times4^n}$? Thanks. – Anonymous Mar 28 '12 at 17:36
You can prove that $4^n \gt 9n^2$ for $n \ge 4$ (try it, using induction). And $\lim_{n\to\infty} 2^{1/n} = 1$ – Aryabhata Mar 28 '12 at 17:39
I'll now try it. How do I prove also that given a>b then $\sqrt[n]a>\sqrt[n]b$? – Anonymous Mar 28 '12 at 17:42
@Anonymous: I don't understand. $a \gt b \gt 0$ implies $a^r \gt b^r$ for any real $r \gt 0$. This has standard proofs in text books which define real numbers etc. – Aryabhata Mar 28 '12 at 17:44
Excellent, thank you very much. I'm now trying to prove it fully, if I'll have further questions I'll write it here. Until then, thanks again! – Anonymous Mar 28 '12 at 17:48

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.

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Do you have a reference for the bigger sledgehammer? – Michael Albanese Sep 11 '12 at 8:37
@MichaelAlbanese See these notes of Pete L. Clark. It's also Theorem 3.37 in Walter Rudin's Principles of Mathematical Analysis. – David Mitra Sep 11 '12 at 14:22

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."

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why $\sqrt[n]{4^n+9n^2}= 4\left(1+\frac{9n^2}{4^n}\right)^{1/n}$? – Anonymous Mar 28 '12 at 18:05
@Anonymous $\sqrt[n]{4^{n}+9n^{2}}=\sqrt{4^{n}\left( 1+9n^{2}/4^{n}\right) }=4\sqrt[n]{ 1+9n^{2}/4^{n}}$ – Américo Tavares Mar 28 '12 at 18:18
did you mean $\sqrt[n]{4^{n}+9n^{2}}=\sqrt[n]{4^{n}\left( 1+9n^{2}/4^{n}\right) }=4\sqrt[n]{ 1+9n^{2}/4^{n}}$ ? – Anonymous Mar 28 '12 at 18:28
@Anonymous I correct the typo: Because $$\sqrt[n]{4^{n}+9n^{2}}=\sqrt[n]{4^{n}\left( 1+9n^{2}/4^{n}\right) }=4\sqrt[n]{ 1+9n^{2}/4^{n}}.$$ – Américo Tavares Mar 28 '12 at 18:29
@Anonymous: Depends how much "calculus" you want to use. Two L'Hospital's Rule. One L'Hospital if rewrite as $(3n/2^n)^2$. Or take third term in binomial expansion of $2^n$, thus $2^n>(n)(n-1)/2$. So $3n/2^n<6/(n-1)$, its square is $<36/(n-1)^2$. – André Nicolas Mar 28 '12 at 21:31

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?
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@André Nicolas: Since the first part is the same as your answer should I delete mine? – Américo Tavares Mar 28 '12 at 18:46
Which ineqaulity do I use on the left and right side of this sequence in order to use the squeeze theorem? – Anonymous Mar 28 '12 at 20:07
@Anonymous: I do not use the squeeze theorem here. I rely on the algebraic maniputation in 1. and on the on the limit in 2., which is proved in the question I link to. – Américo Tavares Mar 28 '12 at 20:31