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

2 Answers
Iván Á.
May 5, 2015

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.

Jim H
May 6, 2015

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

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