Question

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

Answer 1

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

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`