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

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

3. Originally Posted by rgep
We have $\displaystyle \frac{x - \tan 2x}{\sin 2x} = \frac{x}{\sin 2x} - \frac{1}{\cos 2x}$. Now consider the subsequences $\displaystyle x = 2\pi n + \frac\pi2$ and $\displaystyle x = 2\pi n + \pi + \frac\pi2$ to see that there is no limit.
Surly; it is sufficient to consider $\displaystyle x=n.\pi$ for $\displaystyle n=1,2,..$ to show that there can be no finite limit,
and $\displaystyle x=\pi/4+n.\pi$ for $\displaystyle n=1,2,..$ to show that it does not go to $\displaystyle +\infty$?

RonL

4. 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).