1. ## please check for me!!!x goes infinity (cos(2x)-cos(3x))/(x^2)

What is the limit as x goes to infinity of (cos(2x)-cos(3x)) / x^2?

-2 <= cos(2x) - cos(3x) <= 2.

This implies that:

-2/x^2 <= [cos(2x) - cos(3x)]/x^2 <= 2/x^2.

Since -2/x^2 and 2/x^2 --> 0 as x --> infinity, then, by the Squeeze Theorem, [cos(2x) - cos(3x)]/x^2 --> 0 as x --> infinity as it is bounded by -2/x^2 and 2/x^2 and the limits are the same.

is this right??

2. Originally Posted by yunho520
What is the limit as x goes to infinity of (cos(2x)-cos(3x)) / x^2?

-2 <= cos(2x) - cos(3x) <= 2.

This implies that:

-2/x^2 <= [cos(2x) - cos(3x)]/x^2 <= 2/x^2.

Since -2/x^2 and 2/x^2 --> 0 as x --> infinity, then, by the Squeeze Theorem, [cos(2x) - cos(3x)]/x^2 --> 0 as x --> infinity as it is bounded by -2/x^2 and 2/x^2 and the limits are the same.

is this right??

Yes it is...nice!

Tonio