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\[\lim_{x \rightarrow 0}[\frac{ e ^{tanx}-e^x }{ tanx - x }]\]

Mathematics
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so just use l'hopitals
try that and then i can fix if you have more problems
i tried L-Hospital..it is werid...

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Other answers:

d(e^tanx -e^x)/dx e^tanx (sec^2 x) -e^x
d(tanx -x)/dx sec^2 x -1
^ yup.
L'Hospital's, Chain Rule and Quotient Rule! I'd die doing this question O_O
But lim x--->0 sec^2 x-1 =0
Again use L'Hospital's Rule ???
sec^2 x-1 = tan^2 x
try using expressing in this form ... sorry don't have much time http://www.wolframalpha.com/input/?i=lim+x-%3E0+%28e^x-1%29%2Fx
this is interesting |dw:1357557494779:dw|
oh my lord...quotient rule will send chills
using L'Hôpital's rule \[ \large \lim_{x\to0}\frac{e^{\tan x}-e^x}{\tan x-x}= \lim_{x\to0}\frac{e^{\tan x}\sec^2x-e^x}{\sec^2x-1} \] \[ \large =\lim_{x\to0}\frac{e^{\tan x}\sec^4x+e^{\tan x}2\sec^2x\tan x-e^x}{2\sec^2x\tan x} \]
I see some factoring... e^tanx sec^2x
L'H rule wont work here, use expansion
:O
Did nt get u @ghazi
L' Hospital's rule wont be helpful here you will be getting 0/0 form use expansion of functions
oh 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
hmm, i got 1, but i used series.
Can u show it up ?
|dw:1357560889084:dw|
multiply and divide by x tan x.
got that ^ ?
Yeah...!...@sirm3d i also ..like..u see ur Approach ..Plzz
\[e^{\tan x}=1+\frac{\tan x}{1!} + \frac{\tan^2x}{2!}+\cdots\\e^{x}=1+\frac{x}{1!} + \frac{x^2}{2!}+\cdots\\e^{\tan x}-e^x=\frac{\tan x - x}{1!}+\frac{\tan^x - x^2}{2!}+\cdots\]
each numerator has a factor \(\displaystyle \tan x - x\) \[\Large \frac{e^{\tan x}-e^x}{\tan x - x}=\frac{(\tan x - x)(\frac{1}{1!}+\frac{\tan x + x}{2!}+\frac{\tan^x+x \tan x+x^2}{3!}+\cdots)}{\tan x - x}\]
That helps...too.....@sirm3d thxxx
yw.

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