# Help with limit question

Hello can someone help me on this one

Give definition for $$\lim_{k \to \infty} a(k) = 0$$ and use this definition to show $$> \lim_{k \to \infty} \frac{\sin(k)}{\sqrt{k}} = 0 >$$

So far as I've gotten in my answer :

The definition is for all $$\varepsilon > 0$$ there exists an $$M > 0$$, s.t $$k>M$$ implies $$a(k) \to 0$$.

Right? I later use this and start with $$|a(k) - 0| < \varepsilon$$.

Maybe I can do something like $$|\frac{\sin(k)}{\sqrt{k}}*\frac{k}{k} - 0| < \varepsilon = |k/\sqrt{k}|<\varepsilon$$

(since $$\sin(x)/x=1$$)

Am I on the right track with this? And I'm not sure how to continue from here?

Your definition is almost correct. At the end, instead of $$a(k)\to0$$, you should have written $$\bigl\lvert a(k)\bigr\rvert<\varepsilon$$.

Since$$\left\lvert\frac{\sin k}{\sqrt k}\right\rvert\leqslant\frac1{\sqrt k},$$then, for every $$\varepsilon>0$$, take $$N\in\mathbb N$$ such that $$\frac1{\sqrt N}<\varepsilon$$. Then$$k\geqslant N\implies\left\lvert\frac{\sin k}{\sqrt k}\right\rvert\leqslant\frac1{\sqrt k}\leqslant\frac1{\sqrt N}<\varepsilon$$

We need to show that $$\forall \epsilon_1>0$$ $$\exists M_1>0$$ such that $$\forall k\ge M_1$$

$$\left|\frac{\sin k}{\sqrt k}\right|< \epsilon_1$$

and since by definition of limit $$\exists M_2>0$$ such that $$\forall k\ge M_2$$

$$\left|\sin k\right|< \epsilon_2$$

we have also that

$$\left|\frac{\sin k}{\sqrt k}\right|< \left|\frac{\epsilon_2}{\sqrt k}\right|<\epsilon_2$$

then it suffices to assume $$\epsilon_1=\epsilon_2,\quad M_1=M_2$$

to show that $$\frac{\sin k}{\sqrt k}\to 0$$.