How do you find the limit of $x*sin(1/x)$ as x tends to positive infinity?

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Answer 1 · 2015-05-20T15:39:46

You can use L'Hopital's Rule to say the answer is 1:

$lim_{x->infty}xsin(1/x)=lim_{x->infty}\frac{sin(1/x)}{1/x}$

$lim_{x->infty}\frac{-x^{-2}cos(1/x)}{-x^{-2}}=lim_{x->infty}cos(1/x)=cos(0)=1$ .

Answer 2 · 2015-05-20T16:35:14

If you prefer to find the limit without using l'Hopital, rewrite:

$lim_(xrarroo) xsin(1/x) = lim_(xrarroo)(sin(1/x)/(1/x))$

Let $theta = 1/x$ and observe that as $x rarr oo$ , $theta rarr0$ .

So
$lim_(xrarroo) xsin(1/x) = lim_(theta rarr 0)(sin theta/theta) = 1$

How do you find the limit of x*sin(1/x)x*sin(1/x) as x tends to positive infinity?