Determining if the sequence convergers or diverges(III)

[shamieh]

Determine if the sequence converges or diverges, if it converges find the limit

\(\displaystyle n sin\frac{1}{n}\)

so what I did was \(\displaystyle \frac{sin(1/n)}{1/n}\) and then then took the limit as n --> infinity and got 1...Which I guess i really didn/t need to divide by 1/n but oh well.. Would it then be correct to say that the sequence converges to 1 as n--> inifnity? Because I also know that the sin will make it go negative sometimes as well as positive in some cases

 

[Prove It]

Determine if the sequence converges or diverges, if it converges find the limit

\(\displaystyle n sin\frac{1}{n}\)

so what I did was \(\displaystyle \frac{sin(1/n)}{1/n}\) and then then took the limit as n --> infinity and got 1...Which I guess i really didn/t need to divide by 1/n but oh well.. Would it then be correct to say that the sequence converges to 1 as n--> inifnity? Because I also know that the sin will make it go negative sometimes as well as positive in some cases

Well actually you do need to divide by 1/n, because by substituting $\displaystyle \begin{align*} h = \frac{1}{n} \end{align*}$ and noting that as $\displaystyle \begin{align*} n \to \infty , h \to 0 \end{align*}$ you get

$\displaystyle \begin{align*} \lim_{n \to \infty} \frac{\sin{ \left( \frac{1}{n} \right) }}{\frac{1}{n}} &= \lim_{h \to 0} \frac{\sin{(h)}}{h} \\ &= 1 \end{align*}$

you wouldn't be able to do this without dividing by 1/n. But to answer your question, yes, the sequence converges to 1.