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I ran into a block when trying to show that $\lim_{n \to \infty} \frac{\sin n^2}{\sqrt[3] n} = 0$. Using the definition, I wrote,

$$ \left| \frac{\sin n^2}{n^{1/3}} - 0 \right| < \varepsilon \quad \text{for } \varepsilon > 0 $$

subtract zero and notice $n^{1/3} > 0 \quad \forall n \in \mathbb{N}$,

\begin{align*} \frac{ \left| \sin n^2 \right| }{n^{1/3}} &< \varepsilon \\ \left| \sin n^2 \right| &< n^{1/3}\varepsilon \\ \left| \sin n^2 \right|^3 &< n\varepsilon^3 \end{align*}

I know we need to be able to find $n$ given $\varepsilon$. I'm not sure how to isolate $n$ or proceed from here.

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    $\begingroup$ use that $$|\sin(n^2)|\le 1$$ $\endgroup$ Jun 15, 2017 at 17:37
  • $\begingroup$ $n$ is going to $\infty?$ $\endgroup$
    – nouret
    Jun 15, 2017 at 17:37
  • $\begingroup$ @nouret yes indeed $\endgroup$
    – Dando18
    Jun 15, 2017 at 17:38

1 Answer 1

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Note that $|\sin n^2|\leq 1$ for any $n\in \mathbb{N}$. Now $|\sin n^2 /\sqrt[3]n|\leq 1/\sqrt[3]n\leq 1/m_n$, where $m_n$ is largest integral value less than or equals to $\sqrt[3]n$. Now you can always find a $m_n$ for a given value of $\varepsilon >0$ such that $1/m_n\leq \varepsilon$(Why?) which leads to $\lim_{n\to\infty}\frac{\sin n^2}{ \sqrt[3]n}=0$.

Explanation of "Why" part:(Don't see if you want to try yourself!

Suppose does not hold then for all $N\in \mathbb{N}$, $\varepsilon <1/N\Rightarrow N<1/\varepsilon$. But for a fixed $\varepsilon$, $N_{\varepsilon}+1\geq 1/\varepsilon$, and $N_{\varepsilon}$ always exists,(where $N_{\varepsilon}$ is largest integral value less than or equals to $1/{\varepsilon}$), which contradicts our assumption. $\blacksquare$

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