## anonymous 4 years ago limit x->0, tan2x/x find the limit, help!!

1. anonymous

l'Hopital's rule top and bottom. (2sec(x)^2)/1, limx->0 of 2sec(x)^2

2. anonymous

we havent learned that rule, yet, so u wouldnt have tan=cos/sin?

3. anonymous

sin/cos*

4. TuringTest

is it$\tan(2x)$or$\tan^2(x)$?

5. anonymous

tan2x not ^2

6. anonymous

I wonder, I don't think it's possible to solve tan(2x)/x without l'Hopital's. Unless I'm derping. Maybe some weird half angle identity?

7. anonymous

my calc class just started last week n my professor has mentioned that rule but has yet to teach it, so wa does that rule state?

8. anonymous

http://mathworld.wolfram.com/LHospitalsRule.html Apparently I spelled it wrong, but w/e. The proof is lovely.

9. anonymous

you could try $\lim_{x\to 0}\frac{\sin(2x)}{x\cos(2x)}$

10. anonymous

thats as far as i got what do i after that satel.?

11. anonymous

break into to parts $\lim_{x\to 0}\frac{1}{\cos(2x)}= 1$ and $\lim_{x\to 0}\frac{\sin(2x)}{x}=2$

12. TuringTest

How about$\frac{\tan(2x)}x=\frac{\sin(2x)}{x\cos(2x)}=2\frac{\sin x}{x}\cdot\frac{\cos x}{2\cos^2x-1}=2$sat beat me as usual...

13. anonymous

yeah but you are quicker because i did not explain my second limit

14. anonymous

Man, algebra. I forget that if I can't solve a problem, it's probably just me doing algebra badly.

15. Zarkon

I would write $\frac{\sin(2x)}{x}=2\frac{\sin(2x)}{2x}$

16. TuringTest

dang...lol

17. anonymous

or replace x by x/2

18. anonymous

turning test, so 2cos^2x-1 cancels the top, n ur left wit 2?

19. TuringTest

I didn't even need to do that identity, don't know why I did but you can just plug in x=0 to that second part and get 1

20. anonymous

21. TuringTest

that's what you get in total$\frac{\tan(2x)}x=\frac{\sin(2x)}{x\cos(2x)}=2\frac{\sin x}{x}\cdot\frac{\cos x}{\cos(2x)}=2\cdot1\cdot1=2$

22. TuringTest

I didn't write the limits in ...lazy I know

23. anonymous

got it!!