• October 4th 2012, 02:21 PM
ubhutto
how do i prove lim (x^2) * (cos(1/x)) = 0
as x approaches 0
please can you list the steps
• October 4th 2012, 02:49 PM
johnsomeone
Note that cos(1/x) is bounded, no matter how rapidly it oscilates when x is near 0. Then apply the squeeze theorem.
• October 4th 2012, 04:48 PM
Prove It
Quote:

Originally Posted by johnsomeone
Note that cos(1/x) is bounded, no matter how rapidly it oscilates when x is near 0. Then apply the squeeze theorem.

Or just the fact that the product of a function that goes to 0 with a bounded function is 0.
• October 5th 2012, 08:20 AM
ubhutto
$-1\le\cos\left(\tfrac{1}{x}\right)\le 1$ thus $-x^2\le x^2\cos\left(\tfrac{1}{x}\right)\le x^2$
Both $\lim _{x \to 0} - x^2 = 0 = \lim _{x \to 0} x^2$.