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


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