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