How to find the following limit $L=\lim_{n\to\infty}\frac{n^4+n2^n+7}{n^4+3^n+1}?$

Question

How to find the following limit $$L=\lim_{n\to\infty}\dfrac{n^4+n2^n+7}{n^4+3^n+1}?$$ My try: $$L=\lim_{n\to\infty}\dfrac{n^4\left(1+\frac{2^n}{n^3}+\frac{7}{n^4}\right)}{n^4\left(1+\frac{3^n}{n^4}+\frac{1}{n^4}\right)},$$ but I think that's still an indeterminate form $\left[\frac{\infty}{\infty}\right]$. So I tried something else $$L=\lim_{n\to\infty}\dfrac{2^n\left(\frac{n^4}{2^n}+n+\frac{7}{2^n}\right)}{3^n\left(\frac{n^4}{3^n}+1+\frac{1}{3^n}\right)}=\lim_{n\to\infty}\left(\dfrac{2}{3}\right)^nn$$ which is still $[0\cdot\infty]$. How do I solve it?

Answer 1

Option using the binomial expansion:

${\small 0<\dfrac{n}{(3/2)^n}= \dfrac{n}{(1+1/2)^n} <} $

${\small \dfrac{n} {1+n(1/2)+(n(n-1)/2!)(1/2)^2}} $

${\small <\dfrac{(2!)(2^2)n}{n(n-1)}=\dfrac{8}{n-1}}$

Take the limit.

Answer 2

This step

$$L=\lim_{n\to\infty}\dfrac{2^n\left(\frac{n^4}{2^n}+n+\frac{7}{2^n}\right)}{3^n\left(\frac{n^4}{3^n}+1+\frac{1}{3^n}\right)}=\lim_{n\to\infty}\left(\dfrac{2}{3}\right)^nn$$

leads to a correct evaluation but it isn't formally correct.

I suggest to factor out the strongest term from numerator $n2^n$

$$\dfrac{n^4+n2^n+7}{n^4+3^n+1}=\dfrac{n2^n\left(\frac{n^3}{2^n}+1+\frac{7}{n2^n}\right)}{3^n\left(\frac{n^4}{3^n}+1+\frac{1}{3^n}\right)} =\frac{n2^n}{3^n}\cdot \dfrac{\frac{n^3}{2^n}+1+\frac{7}{n2^n}}{\frac{n^4}{3^n}+1+\frac{1}{3^n}} \to 0\cdot 1 =0$$

for example using that, since $\frac{\log n}n \to 0$

$$\frac{n2^n}{3^n}=e^{\log n-n\log \frac32}=e^{n\left(\frac{\log n}n-\log \frac32\right)} \to e^{-\infty}$$

or by ratio or root test.

Answer 3

I would simply use the sandwich theorem.

Clearly the sequence are all nonnegative numbers.

So the resulting limit is at least $0$ if it converges.

$\frac{n^4 + n 2^n + 7}{n^4 + 3^n + 1} < \frac{n^4 + n 2^n + 7}{3^n} = \frac{n^4}{3^n} + \frac{n 2^n}{3^n} + \frac{7}{3^n}$

So

$0 =< \lim_{n \to +\infty} \frac{n^4 + n 2^n + 7}{n^4 + 3^n + 1} < \frac{n^4 + n 2^n + 7}{3^n} = \lim_{n \to +\infty} \frac{n^4}{3^n} + \lim_{n \to +\infty} \frac{n 2^n}{3^n} + \lim_{n \to +\infty} \frac{7}{3^n} =< 0 + 0 + 0 =< 0$

So the limit is $0$.

Answer 4

A hint:You can see $$a(n)=n(\frac 23)^n$$ for $n\ge 3$ is decreasing ,also positive. This mean $$n \ge 3 \to 0< \cdots <a(n+1)<a(n)<a(n-1)<\cdots<a(3)$$ and finally $n(\frac 23)^n \to 0 $

Answer 5

Let $a_n=n(\frac{2}{3})^n$. You know that if $\lim\limits_{n\to\infty} \sqrt[n]{a_n}<1$ then $\lim\limits_{n\to\infty} a_n=0$.
$\lim\limits_{n\to\infty} \sqrt[n]{n(\frac 23)^n}=\lim\limits_{n\to\infty}\sqrt[n]{n} \cdot \frac{2}{3}\underset{\ast}{=} \frac 23<1.$

$\ast$ - Limit of the sequence $\lim_{n\rightarrow\infty}\sqrt[n]n$