Question

find the limit as it approaches infinity

(cos2x / 3x).

I thought it'd be 2/3 but my answer key says 0. I think cos2x can be rewrote as cos^2x - sin^2x but that just confuses me more?

Answer 1

use Squeeze Theorem: http://en.wikipedia.org/wiki/Squeeze_theorem

\[\frac{ \cos(2x) }{ 3x } = \frac{ 1 }{ 3 }\frac{ \cos(2x) }{ x }\]

and note that

\[-1 \le \cos(2x) \le 1 => \frac{ -1 }{ x } \le \frac{ \cos(2x) }{ x } \le \frac{ 1 }{ x }\]

since lim approaching inf of 1/x and -1/x is zero, by squeeze theorem, limit of cos(2x)/x = 0

Answer 2

the lazy approach is just recognize that cos is restricted to values between -1 and 1
thus cos(infinity) = constant
lim = constant/infinity = 0

Answer 3

cos2x is bounded so bounded/infinity =0