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Yahoo!

  • 3 years ago

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

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  1. RONNCC
    • 3 years ago
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    so just use l'hopitals

  2. RONNCC
    • 3 years ago
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    try that and then i can fix if you have more problems

  3. Yahoo!
    • 3 years ago
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    i tried L-Hospital..it is werid...

  4. sauravshakya
    • 3 years ago
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    d(e^tanx -e^x)/dx e^tanx (sec^2 x) -e^x

  5. sauravshakya
    • 3 years ago
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    d(tanx -x)/dx sec^2 x -1

  6. RONNCC
    • 3 years ago
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    ^ yup.

  7. ParthKohli
    • 3 years ago
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    L'Hospital's, Chain Rule and Quotient Rule! I'd die doing this question O_O

  8. sauravshakya
    • 3 years ago
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    But lim x--->0 sec^2 x-1 =0

  9. sauravshakya
    • 3 years ago
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    Again use L'Hospital's Rule ???

  10. Yahoo!
    • 3 years ago
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    sec^2 x-1 = tan^2 x

  11. experimentX
    • 3 years ago
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    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
    • 3 years ago
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    this is interesting |dw:1357557494779:dw|

  13. Yahoo!
    • 3 years ago
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    @sauravshakya @experimentX

  14. UsukiDoll
    • 3 years ago
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    oh my lord...quotient rule will send chills

  15. helder_edwin
    • 3 years ago
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    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
    • 3 years ago
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    I see some factoring... e^tanx sec^2x

  17. ghazi
    • 3 years ago
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    L'H rule wont work here, use expansion

  18. UsukiDoll
    • 3 years ago
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    :O

  19. Yahoo!
    • 3 years ago
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    Did nt get u @ghazi

  20. ghazi
    • 3 years ago
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    L' Hospital's rule wont be helpful here you will be getting 0/0 form use expansion of functions

  21. UsukiDoll
    • 3 years ago
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    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
    • 3 years ago
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    hmm, i got 1, but i used series.

  23. Yahoo!
    • 3 years ago
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    Can u show it up ?

  24. hartnn
    • 3 years ago
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    |dw:1357560889084:dw|

  25. hartnn
    • 3 years ago
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    multiply and divide by x tan x.

  26. hartnn
    • 3 years ago
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    got that ^ ?

  27. Yahoo!
    • 3 years ago
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    Yeah...!...@sirm3d i also ..like..u see ur Approach ..Plzz

  28. sirm3d
    • 3 years ago
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    \[e^{\tan x}=1+\frac{\tan x}{1!} + \frac{\tan^2x}{2!}+\cdots\\e^{x}=1+\frac{x}{1!} + \frac{x^2}{2!}+\cdots\\e^{\tan x}-e^x=\frac{\tan x - x}{1!}+\frac{\tan^x - x^2}{2!}+\cdots\]

  29. sirm3d
    • 3 years ago
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    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!
    • 3 years ago
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    That helps...too.....@sirm3d thxxx

  31. sirm3d
    • 3 years ago
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    yw.

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