Question about calculating $\lim\limits_{n \to\infty} \dfrac{6n^5+(\sin(8n))^2}{n^4+6} $

Question

I have the following limit:

$$\lim\limits_{n \to\infty} \frac{6n^5+(\sin(8n))^2}{n^4+6} $$

My fist question is, can I solve this by dividing by the highest n power? The reason I was uncertain about doing this was because of the $(\sin(8n))^2 $ however, I figured as it is always bounded by 0 and 1, dividing by n to the 5th power would cause it to tend to 0. Is this wrong?

My second question is, if I can solve this by dividing by the highest n power, I get $\frac{6}{0} $ so how do I know whether this tends to positive or negative infinity? The question did not specify whether n was originally tending to positive or negative infinity.

Thank you!

Your first question: Your reasoning is good. Your second question: Just look at it. Is it ever negative? — Simply Beautiful Art, Jan 07 '17 at 22:42
@MathChat He is wondering if the limit itself tends to $+/-\infty$. — Simply Beautiful Art, Jan 07 '17 at 22:47
@ Simple Art: I'm not sure, because of his comment :" The question did not specify whether n was originally tending to positive or negative infinity." — math, Jan 07 '17 at 22:49

score 1 · Accepted Answer · answered Jan 07 '17 at 22:47

For your first question, we can rewrite as follows:

$$\frac{6n^5+\sin^2(8n)}{n^4+6}=\frac{6n^5}{n^4+6}+\frac{\sin^2(8n)}{n^4+6}$$

The first fraction is the real problem, but the second fraction can easily be handled with squeeze theorem for the reason you gave $\sin(x)$ shouldn't be a problem here.

Second question:

Upon then dividing by highest powers, you get $\frac60$, and you want to know if the result is positive or negative infinity. This may be done by noticing that

$$n^4+6\ge6\\6n^5>0;\quad n>0\\\sin^2(8n)\ge0$$

Since anything squared is positive, and I don't think I need to give much explanation as to why $6n^5$ will be positive.

Putting all that together, you should be able to conclude it diverges to $+\infty$.

Brevan Ellefsen · Answer 2 · 2017-01-07T23:06:37.027

1

1) You are not wrong here. $\sin(ax)$ is always bounded between $0$ and $1$, and when you square any value in $[0,1]$ you get another value in $[0,1]$. We thus don't have to worry about the sine function exceeding this bound.

2) It doesn't matter if $n\to+\infty$ or $n\to-\infty$, as both will give you the same answer (although $n\to\infty$ usually means $n \to +\infty$). Note that all the time $n$ shows up it is raised to an even power, meaning that it doesn't matter if $n$ is negative or positive. Symbolically, this means that $(-n)^{2x} = (-n)^{2^x} = \left((-1)^2\right)^x n^{2x}= 1^xn^{2x} = n^{2x}$

As far as actually investigating your limit goes, I would first divide by $n^4$. We get that $$\lim\limits_{n \to\infty} \frac{6n^5+(\sin(8n))^2}{n^4+6}$$ $$=\lim\limits_{n \to\infty} \frac{6n+\frac{(\sin(8n))^2}{n^4}}{1+\frac{6}{n^4}}$$ As you noted, the sine is bounded, and so dividing by an increasing $n$ will make the term tend to zero. Likewise, $\lim_{n\to\infty} \frac{6}{n^4} = 0$. Applying these, we get that the limit is just $$\lim\limits_{n \to\infty} 6n$$
Which clearly diverges to $+\infty$. I am not sure how to got $\frac{6}{0}$ and not $\frac{6}{1}$, though I would be happy to investigate your work if you post it!

edited Jan 07 '17 at 23:06

answered Jan 07 '17 at 22:51

Brevan Ellefsen

14,793

You stated you would divide by n to the 5th power and you then divided by n to the 4th power. Why? – melm Jan 07 '17 at 23:00
@aa21 oops, my bad. I meant the 4th power. I'll fix that. If you divide by the 5th power you get $$=\lim\limits_{n \to\infty} \frac{6+\frac{(\sin(8n))^2}{n^5}}{\frac{1}{n}+\frac{6}{n^5}}$$. The problem now is that, when you have a fixed numerator on a fraction and the denominator shrinks, then the fraction gets bigger. Take $\lim_{n\to \infty} \frac{6}{\frac{1}{n}} = \infty$ for example. I divided by $n^4$ because it makes it clearer that the function is growing – Brevan Ellefsen Jan 07 '17 at 23:06
Oh, so I can still divide by n to the 5th power and get the same answer as when I would divide by the 4th power? – melm Jan 07 '17 at 23:08
IMO, $6/0$ is also pretty clear as $\pm\infty$ limit. Though maybe not the sign I suppose. – Simply Beautiful Art Jan 07 '17 at 23:08
@aa21 Any correct algebraic manipulation shall not change the end result of a limit. – Simply Beautiful Art Jan 07 '17 at 23:08
@SimpleArt Thank you!! I get it now :) – melm Jan 07 '17 at 23:10
@aa21 Your welcome $\ddot\smile$ – Simply Beautiful Art Jan 07 '17 at 23:10
@SimpleArt *You're. How dare you fill my answer's comment section with grammatical mistakes :0 – Brevan Ellefsen Jan 08 '17 at 01:42
:O this is terrible! I must eradicate all evidence...including you... – Simply Beautiful Art Jan 08 '17 at 01:46
@SimpleArt *hides in the broom closed* – Brevan Ellefsen Jan 08 '17 at 01:47

score 0 · Answer 3 · answered Jan 07 '17 at 22:51

0

$\lim\limits_{n \to\infty} \frac{6n^5+\sin^28n}{n^4+6}=\lim\limits_{n \to\infty} \frac{6n+\frac{\sin^28n}{n^4}}{1+\frac{6}{n^4}} =\infty $

answered Jan 07 '17 at 22:51

Michael Rozenberg

194,933

Why did you divide by n to the 4th power? I thought we could only divide by the highest n power? – melm Jan 07 '17 at 23:01
@aa21 If so we'll get $0$ in the denominator. It's better to divide by $n^4$. – Michael Rozenberg Jan 07 '17 at 23:19

score 0 · Answer 4 · answered Jan 07 '17 at 22:54

0

It is much simpler to use equivalents:

As $\sin^28n$ is bounded, $6n^5+\sin^28n\sim_\infty 6n^5$; $n^4+6\sim_\infty n^4$, so $$\frac{6n^5+\sin^28n}{n^4+6}\sim_\infty\frac{6n^5}{n^4}=6n,$$ and the limit is $+\infty$.

answered Jan 07 '17 at 22:54

Bernard

175,478

Question about calculating $\lim\limits_{n \to\infty} \dfrac{6n^5+(\sin(8n))^2}{n^4+6} $

4 Answers4