How to use ε-N language to prove $\lim_{n\to ∞}\frac{\sqrt[3]{n^2}\sin n}{n+1}=0$

Question

The statement

$$\lim_{n\to ∞}\frac{\sqrt[3]{n^2}\sin n}{n+1}=0$$ is equivalent to

$$\forall \epsilon > 0 \quad \exists N_\epsilon \quad \mathrm{s.t.} \quad \forall n \geq N_\epsilon \quad \left|\frac{\sqrt[3]{n^2}\sin n}{n+1}-0 \right|\leq \epsilon.$$

I don't know how to express $N_\epsilon$ in terms of $\epsilon$ in this question.

Answer 1

Notice that, for all $n$, because $|\sin n| \leq 1$ and $1/|n+1| \leq 1/|n|$, $$|f(n)| =\left|\frac{n^{2/3} \sin n}{n+1}\right| \leq \left|\frac{n^{2/3}}{n+1}\right| \leq \left| \frac{n^{2/3}}{n}\right| = \left| \frac 1 {n^{1/3}} \right|. $$ So $|f(n)| \leq \epsilon$ when $\left|n^{-1/3} \right|\leq \epsilon$, that is, when $$n \geq \frac 1 {\epsilon^3} =: N_\epsilon.$$

Answer 2

Try to decompose the formula and see how the components behave individually as $n \to \infty$. How does $\sqrt[3]{n^2}$ behave? What about $\frac{1}{n+1}$? What effect do the oscillations of $\sin n$ have on the terms of the sequence (what is the limited range of $\sin n$)?