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

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.

## closed as unclear what you're asking by qbert, user99914, Namaste, Ethan Bolker, José Carlos SantosJul 7 '18 at 20:35

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

• There's no $x$ in your formula.... Do you mean $\lim_{n\to\infty}\cdots$? – Lord Shark the Unknown Jul 7 '18 at 2:11
• @LordSharktheUnknown sorry, I made the mistake when I paste the code.... – Chloe Zhou Jul 7 '18 at 2:14
• Can you use squeeze theorem? That would be a lot easier. – user122049 Jul 7 '18 at 2:19

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 $$n \geq \frac 1 {\epsilon^3} =: N_\epsilon.$$
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 $\sin n$)?
• @ChloeZhou Yes! But before you dive into finding the $n$ for your $\varepsilon$, try to think how the sequence behaves in terms of the components. Why do the terms get arbitrarily close to $0$? – Andrey Portnoy Jul 7 '18 at 2:19