A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 4 years ago
limit x>0, tan2x/x find the limit, help!!
anonymous
 4 years ago
limit x>0, tan2x/x find the limit, help!!

This Question is Closed

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0l'Hopital's rule top and bottom. (2sec(x)^2)/1, limx>0 of 2sec(x)^2

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0we havent learned that rule, yet, so u wouldnt have tan=cos/sin?

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.4is it\[\tan(2x)\]or\[\tan^2(x)\]?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I 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?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0my 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?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0http://mathworld.wolfram.com/LHospitalsRule.html Apparently I spelled it wrong, but w/e. The proof is lovely.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0you could try \[\lim_{x\to 0}\frac{\sin(2x)}{x\cos(2x)}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0thats as far as i got what do i after that satel.?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0break into to parts \[\lim_{x\to 0}\frac{1}{\cos(2x)}= 1\] and \[\lim_{x\to 0}\frac{\sin(2x)}{x}=2\]

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.4How about\[\frac{\tan(2x)}x=\frac{\sin(2x)}{x\cos(2x)}=2\frac{\sin x}{x}\cdot\frac{\cos x}{2\cos^2x1}=2\]sat beat me as usual...

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yeah but you are quicker because i did not explain my second limit

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Man, algebra. I forget that if I can't solve a problem, it's probably just me doing algebra badly.

Zarkon
 4 years ago
Best ResponseYou've already chosen the best response.1I would write \[\frac{\sin(2x)}{x}=2\frac{\sin(2x)}{2x}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0turning test, so 2cos^2x1 cancels the top, n ur left wit 2?

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.4I 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

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.4that's what you get in total\[\frac{\tan(2x)}x=\frac{\sin(2x)}{x\cos(2x)}=2\frac{\sin x}{x}\cdot\frac{\cos x}{\cos(2x)}=2\cdot1\cdot1=2\]

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.4I didn't write the limits in ...lazy I know
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.