A community for students.
Here's the question you clicked on:
 0 viewing
xilinix
 3 years ago
lim > 0 sin(2x) /(x*cos(x) )
help
xilinix
 3 years ago
lim > 0 sin(2x) /(x*cos(x) ) help

This Question is Closed

myko
 3 years ago
Best ResponseYou've already chosen the best response.0sin (2x) = 2sin(x)cos(x) so sin(2x) /(x*cos(x) =2sin(x)cos(x)/x*cos(x)=2sinx/x which as x>0 goes to 1

myko
 3 years ago
Best ResponseYou've already chosen the best response.0ups sry, forgot about 2 multuplying: lim 2sinx/x=2*1=2 x>0

myko
 3 years ago
Best ResponseYou've already chosen the best response.0sin (2x) can be written like: sin (2x) = 2sin(x)cos(x) substituting this expretion in the original one: sin(2x) /(x*cos(x) =2sin(x)cos(x)/x*cos(x)=2sinx/x taking limits, and noticing that lim as x>0 of sinx/x=1: lim 2sinx/x=2*1=2 x>0

xilinix
 3 years ago
Best ResponseYou've already chosen the best response.1why not? l'Hôpital rule

ccappiello
 3 years ago
Best ResponseYou've already chosen the best response.0yes, using l'hospital's rule works

Varuns
 3 years ago
Best ResponseYou've already chosen the best response.0Don't break your head ! You know that sin(x)=x as x tends to zero. So sin(2x) = 2x. In the denominator, keep x as it is and cos(x)=1. Hence the answer

xilinix
 3 years ago
Best ResponseYou've already chosen the best response.1thanks, but the image is ok?

lopus
 3 years ago
Best ResponseYou've already chosen the best response.1claro que es posible es solamente derivar arriba y abajo como lo muestras en la imagen y ya
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.