I feel like I have to find a way to cancel out the $n$'s but I'm not sure how to do so with the $\sin^2(11n)$ The limit is from $n$ to $\infty$, I tried dividing everything by $n^3$ but I don't know how to deal with that
$$ \frac{-4n^3+\sin^2(11n)}{n^3+17} $$
 A: Using the squeeze theorem, for example:
$$-4\xleftarrow[n\to\infty]{}\frac{-4n^3}{n^3+17}\le\color{red}{\frac{-4n^2+\sin^211n}{n^3+17}}\le\frac{-4n^3+1}{n^3}=-4+\frac1{n^3}\xrightarrow[n\to\infty]{}-4$$
A: The limit is $-4$.
$$\lim_{n \to \infty} \frac{-4n^3+\sin^2(11n)}{n^3+17}$$
If $n$ approaches infinity, the  $+\sin^2(11n)$ in the nominator (It is bounded by $[0,1]$) and $+17$ in the denominator do not matter anymore. Therefore the limit is the same as $$\lim_{n \to \infty} \frac{-4n^3}{n^3}$$ Which is $-4$.
A: You can try splitting up the fraction and then multiplying through one of the terms by $\frac{n^{-3}}{n^{-3}}$. $$\begin{align} \frac{-4n^3+\sin^2(11n)}{n^3+17} =\frac{\sin^2(11n)}{n^3+17} -\frac{4n^3}{n^3+17}\cdot\frac{n^{-3}}{n^{-3}}  \\ =\frac{\sin^2(11n)}{n^3+17} -\frac{4}{1+\frac{17}{n^3}} \end{align}$$ You should be able to see that $\frac{\sin^2(11n)}{n^3+17}$ converges to zero as $n \to \infty$ because the numerator $\sin^2(11n)$ is bounded between $0$ and $1$, hence the denominator will grow exponentially while the numerator is bounded. Then the term $-\frac{4}{1+\frac{17}{n^3}}$ is nice because  $\frac{17}{n^3}$ converges to zero as $n \to \infty$, so $-\frac{4}{1+\frac{17}{n^3}} \to -\frac{4}{1+0} = -4$. In conclusion, $$\lim_{n \to \infty} \frac{-4n^3+\sin^2(11n)}{n^3+17}  = \lim_{n \to \infty}\frac{\sin^2(11n)}{n^3+17} -\frac{4}{1+\frac{17}{n^3}}  = -4 $$
