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

Question

William Abdalla · Accepted Answer

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.

William Abdalla · Answer

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#

Scarlett Allison · Answer

undefined To find the limit of x^2cos(pi / x) as x approaches 0, we can use the squeeze theorem.

First, we note that -1 ≤ cos(pi / x) ≤ 1 for all x.

Next, we multiply both sides of the inequality by x^2 (since x^2 is always non-negative) to get:

-x^2 ≤ x^2cos(pi / x) ≤ x^2

As x approaches 0, both -x^2 and x^2 approach 0.

Therefore, by the squeeze theorem, the limit of x^2cos(pi / x) as x approaches 0 is 0.