# Limits

• December 30th 2005, 07:53 PM
bobby77
Find the trigonometric limit: lim (as x approaches infinity) (x-tan 2x)/sin 2x
• December 30th 2005, 11:10 PM
rgep
We have $\frac{x - \tan 2x}{\sin 2x} = \frac{x}{\sin 2x} - \frac{1}{\cos 2x}$. Now consider the subsequences $x = 2\pi n + \frac\pi2$ and $x = 2\pi n + \pi + \frac\pi2$ to see that there is no limit.
• December 30th 2005, 11:29 PM
CaptainBlack
Quote:

Originally Posted by rgep
We have $\frac{x - \tan 2x}{\sin 2x} = \frac{x}{\sin 2x} - \frac{1}{\cos 2x}$. Now consider the subsequences $x = 2\pi n + \frac\pi2$ and $x = 2\pi n + \pi + \frac\pi2$ to see that there is no limit.

Surly; it is sufficient to consider $x=n.\pi$ for $n=1,2,..$ to show that there can be no finite limit,
and $x=\pi/4+n.\pi$ for $n=1,2,..$ to show that it does not go to $+\infty$?

RonL
• December 31st 2005, 10:01 PM
ticbol
Quote:

Originally Posted by bobby77
Find the trigonometric limit: lim (as x approaches infinity) (x-tan 2x)/sin 2x

Are you sure that it is not as x approaches zero?

If it is as x approaches infinity, then what is tan(infinity)? Or, sin(infinity)?

If x approaches zero, limit is (-1/2).