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