Question

How do you find the limit of `xsin(pi/x)` as x approaches infinity?

Answer 1

Use `lim_(thetararr0)sintheta/theta` and some algebra.

Explanation:

`xsin(pi/x) = sin(pi/x)/(1/x)`

`= pi/pi sin(pi/x)/(1/x)`

`= pi sin(pi/x)/(pi/x)`

With `theta = pi/x`, we have

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

`= pi lim_(xrarroo) sin(pi/x)/(pi/x)`

`= pi lim_(thetararr0) sintheta/theta`

`= pi`