How can prove limit of $ n\sin(\frac{1}{n})=1 $ as $ n $ approaches to infinity?

Asked Aug 06 '22 at 15:23

Active Aug 06 '22 at 15:33

Viewed 47 times

I am trying to solve this sequence limit

$\lim_{n\to \infty} $ $n\sin(\frac{1}{n})=1$

and I’m struggling finding a bound for $|a_n -l|$

I thought of rewriting

$|n\sin(\frac{1}{n}) - 1|$

$|\frac {\sin(\frac{1}{n})}{\frac{1}{n}} - 1|$

An then maybe rewrite 1 as

$|\frac {\sin(\frac{1}{n})}{\frac{1}{n}} - \frac{\frac{1}{n}}{\frac{1}{n}}|$

But I don’t know how to continue and find $\epsilon$

I would appreciate some help and thanks in advance!

edited Aug 06 '22 at 15:33

Sourav Ghosh

asked Aug 06 '22 at 15:23

santijrv

You could use L'Hopital for $\sin(x)/x$. – Michael Aug 06 '22 at 15:27
There are many characterizations of the sine function (e.g., power series representation, the inverse of $f(x)=\int_0^x \frac{1}{\sqrt{1-t^2}},dt$. How do you define $\sin(x)$? See THIS – Mark Viola Aug 06 '22 at 15:28
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to. – Sourav Ghosh Aug 06 '22 at 15:35

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