## Yahoo! 2 years ago $\lim_{x \rightarrow 0}[\frac{ e ^{tanx}-e^x }{ tanx - x }]$

1. RONNCC

so just use l'hopitals

2. RONNCC

try that and then i can fix if you have more problems

3. Yahoo!

i tried L-Hospital..it is werid...

4. sauravshakya

d(e^tanx -e^x)/dx e^tanx (sec^2 x) -e^x

5. sauravshakya

d(tanx -x)/dx sec^2 x -1

6. RONNCC

^ yup.

7. ParthKohli

L'Hospital's, Chain Rule and Quotient Rule! I'd die doing this question O_O

8. sauravshakya

But lim x--->0 sec^2 x-1 =0

9. sauravshakya

Again use L'Hospital's Rule ???

10. Yahoo!

sec^2 x-1 = tan^2 x

11. experimentX

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

12. experimentX

this is interesting |dw:1357557494779:dw|

13. Yahoo!

@sauravshakya @experimentX

14. UsukiDoll

oh my lord...quotient rule will send chills

15. helder_edwin

using L'Hôpital's rule $\large \lim_{x\to0}\frac{e^{\tan x}-e^x}{\tan x-x}= \lim_{x\to0}\frac{e^{\tan x}\sec^2x-e^x}{\sec^2x-1}$ $\large =\lim_{x\to0}\frac{e^{\tan x}\sec^4x+e^{\tan x}2\sec^2x\tan x-e^x}{2\sec^2x\tan x}$

16. UsukiDoll

I see some factoring... e^tanx sec^2x

17. ghazi

L'H rule wont work here, use expansion

18. UsukiDoll

:O

19. Yahoo!

Did nt get u @ghazi

20. ghazi

L' Hospital's rule wont be helpful here you will be getting 0/0 form use expansion of functions

21. UsukiDoll

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

22. sirm3d

hmm, i got 1, but i used series.

23. Yahoo!

Can u show it up ?

24. hartnn

|dw:1357560889084:dw|

25. hartnn

multiply and divide by x tan x.

26. hartnn

got that ^ ?

27. Yahoo!

Yeah...!...@sirm3d i also ..like..u see ur Approach ..Plzz

28. sirm3d

$e^{\tan x}=1+\frac{\tan x}{1!} + \frac{\tan^2x}{2!}+\cdots\\e^{x}=1+\frac{x}{1!} + \frac{x^2}{2!}+\cdots\\e^{\tan x}-e^x=\frac{\tan x - x}{1!}+\frac{\tan^x - x^2}{2!}+\cdots$

29. sirm3d

each numerator has a factor $$\displaystyle \tan x - x$$ $\Large \frac{e^{\tan x}-e^x}{\tan x - x}=\frac{(\tan x - x)(\frac{1}{1!}+\frac{\tan x + x}{2!}+\frac{\tan^x+x \tan x+x^2}{3!}+\cdots)}{\tan x - x}$

30. Yahoo!

That helps...too.....@sirm3d thxxx

31. sirm3d

yw.