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Use the fact that $\forall x \in \mathbb{R}$, $|\sin(x)| \leq |x|$ to prove that $\{\sin(1/n)\}$ from $n=1$ to $\infty$ converges.

Attempt:

Fix $\epsilon > 0$. Let $N = \lceil\frac{1}{\arcsin(\epsilon)}\rceil$. Then if $n \geq N$,

We have:

$$\left|\sin\left(\frac1n\right) - 0\right| = \sin\left(\frac1n\right) \leq \frac1n \leq \frac1N < \epsilon$$

My attempt at the solution is basically based on another problem. Although, I'm pretty sure it's incorrect. What is the correct way to approach this problem?

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    What is wrong with your solution?! – 2017-02-09
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    @arberavdullahu This problem was based on another problem that I had done. The difference was that in the question it asks me to use the fact |sin(x)| <= |x|. I didn't exactly use this in the process of writing my proof. I basically used the inequality of |sin(1/n) - 0| < epsilon and rearranged it to get n < 1/arcsin(epsilon) and used that for my N. Basically, I didn't know if my proof met all of the requirements. Is my proof a sufficient one? – 2017-02-09

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Let $\epsilon>0$
Since $\frac{1}{n}$ converges to 0 then there exist a $N\in \mathbb{N}$ such that for$\epsilon$ and for $n\geq N$ then $$|\frac{1}{n}|<\epsilon$$ Now $$|sin(\frac{1}{n})-0|\leq |\frac{1}{n}|<\epsilon$$ So here I used that $|sinx|\leq |x|$