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anonymous
 4 years ago
\[\lim_{x \rightarrow 0}[\frac{ e ^{tanx}e^x }{ tanx  x }]\]
anonymous
 4 years ago
\[\lim_{x \rightarrow 0}[\frac{ e ^{tanx}e^x }{ tanx  x }]\]

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so just use l'hopitals

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i tried LHospital..it is werid...

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

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

ParthKohli
 4 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

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

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

experimentX
 4 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
 4 years ago
Best ResponseYou've already chosen the best response.1this is interesting dw:1357557494779:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@sauravshakya @experimentX

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

helder_edwin
 4 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
 4 years ago
Best ResponseYou've already chosen the best response.0I see some factoring... e^tanx sec^2x

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

Ghazi
 4 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
 4 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
 4 years ago
Best ResponseYou've already chosen the best response.2hmm, i got 1, but i used series.

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

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

sirm3d
 4 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
 4 years ago
Best ResponseYou've already chosen the best response.2each numerator has a factor \(\displaystyle \tan x  x\) \[\Large \frac{e^{\tan x}e^x}{\tan x  x}=\frac{(\tan x  x)(\frac{1}{1!}+\frac{\tan x + x}{2!}+\frac{\tan^x+x \tan x+x^2}{3!}+\cdots)}{\tan x  x}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0That helps...too.....@sirm3d thxxx
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