Question

How do you find the limit of `x^2cos(pi / x)` as x approaches 0?

Answer 1

cos(1/x) oscillates between 1 and -1 really fast when aproaching to 0 but it always is a finite number. If you are multipliying this oscillating function to x^2 which modullates the oscillation it will oscillate between x^2 and -x^2.
So if you make the limit going to 0 it will "oscillating" between +0 and -0 which you see is 0.

Answer 2

The limit is `0`.

To show this, use the squeeze (pinch, sandwich) theorem.

`-1<= cos (pi/x)<= 1` for `x!=0`

Because `x^2 >0` for all `x!=0`, we can multiply the inequality without reversing it:

`-x^2<= x^2cos (pi/x)<= x^2` for `x!=0`.

Note that

`lim_(xrarr0)(-x^2) in=0=lim_(xrarr0)x^2`,

So the squeeze theorem (or whatever you call it) tells us that

`lim_(xrarr0) x^2cos (pi/x) =0`