A community for students.
Here's the question you clicked on:
 0 viewing
Yahoo!
 2 years ago
\[\lim_{x \rightarrow 0}[\frac{ e ^{tanx}e^x }{ tanx  x }]\]
Yahoo!
 2 years ago
\[\lim_{x \rightarrow 0}[\frac{ e ^{tanx}e^x }{ tanx  x }]\]

This Question is Closed

RONNCC
 2 years ago
Best ResponseYou've already chosen the best response.0try that and then i can fix if you have more problems

Yahoo!
 2 years ago
Best ResponseYou've already chosen the best response.0i tried LHospital..it is werid...

sauravshakya
 2 years ago
Best ResponseYou've already chosen the best response.0d(e^tanx e^x)/dx e^tanx (sec^2 x) e^x

sauravshakya
 2 years ago
Best ResponseYou've already chosen the best response.0d(tanx x)/dx sec^2 x 1

ParthKohli
 2 years ago
Best ResponseYou've already chosen the best response.0L'Hospital's, Chain Rule and Quotient Rule! I'd die doing this question O_O

sauravshakya
 2 years ago
Best ResponseYou've already chosen the best response.0But lim x>0 sec^2 x1 =0

sauravshakya
 2 years ago
Best ResponseYou've already chosen the best response.0Again use L'Hospital's Rule ???

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.1try using expressing in this form ... sorry don't have much time http://www.wolframalpha.com/input/?i=lim+x%3E0+%28e^x1%29%2Fx

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.1this is interesting dw:1357557494779:dw

Yahoo!
 2 years ago
Best ResponseYou've already chosen the best response.0@sauravshakya @experimentX

UsukiDoll
 2 years ago
Best ResponseYou've already chosen the best response.0oh my lord...quotient rule will send chills

helder_edwin
 2 years ago
Best ResponseYou've already chosen the best response.0using L'Hôpital's rule \[ \large \lim_{x\to0}\frac{e^{\tan x}e^x}{\tan xx}= \lim_{x\to0}\frac{e^{\tan x}\sec^2xe^x}{\sec^2x1} \] \[ \large =\lim_{x\to0}\frac{e^{\tan x}\sec^4x+e^{\tan x}2\sec^2x\tan xe^x}{2\sec^2x\tan x} \]

UsukiDoll
 2 years ago
Best ResponseYou've already chosen the best response.0I see some factoring... e^tanx sec^2x

ghazi
 2 years ago
Best ResponseYou've already chosen the best response.0L'H rule wont work here, use expansion

ghazi
 2 years ago
Best ResponseYou've already chosen the best response.0L' Hospital's rule wont be helpful here you will be getting 0/0 form use expansion of functions

UsukiDoll
 2 years ago
Best ResponseYou've already chosen the best response.0oh wikipedia the true online source for Britannica related stuff, share with me this...KNOWLEDGE! http://en.wikipedia.org/wiki/L%27H%C3%B4pital%27s_rule

sirm3d
 2 years ago
Best ResponseYou've already chosen the best response.2hmm, i got 1, but i used series.

hartnn
 2 years ago
Best ResponseYou've already chosen the best response.1multiply and divide by x tan x.

Yahoo!
 2 years ago
Best ResponseYou've already chosen the best response.0Yeah...!...@sirm3d i also ..like..u see ur Approach ..Plzz

sirm3d
 2 years ago
Best ResponseYou've already chosen the best response.2\[e^{\tan x}=1+\frac{\tan x}{1!} + \frac{\tan^2x}{2!}+\cdots\\e^{x}=1+\frac{x}{1!} + \frac{x^2}{2!}+\cdots\\e^{\tan x}e^x=\frac{\tan x  x}{1!}+\frac{\tan^x  x^2}{2!}+\cdots\]

sirm3d
 2 years ago
Best ResponseYou've already chosen the best response.2each numerator has a factor \(\displaystyle \tan x  x\) \[\Large \frac{e^{\tan x}e^x}{\tan x  x}=\frac{(\tan x  x)(\frac{1}{1!}+\frac{\tan x + x}{2!}+\frac{\tan^x+x \tan x+x^2}{3!}+\cdots)}{\tan x  x}\]

Yahoo!
 2 years ago
Best ResponseYou've already chosen the best response.0That helps...too.....@sirm3d thxxx
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.