Question

What is the limit of `xsin(1/x)` as x goes to infinity?

Answer 1

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

Explanation:

Use `lim_(thetararr0)sintheta/theta = 1` and

`lim_(xrarroo)1/x = 0`