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DLS

  • 3 years ago

limits

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  1. DLS
    • 3 years ago
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    \[\LARGE \lim_{x \rightarrow 0} \frac{1+e^{\frac{-1}{x}}}{1-e^{\frac{-1}{x}}} \]

  2. DLS
    • 3 years ago
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    1) 1 2)-1 3)0 4)Does not exist

  3. hartnn
    • 3 years ago
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    can u use L'Hopitals ?

  4. DLS
    • 3 years ago
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    anything

  5. hartnn
    • 3 years ago
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    then try, x= -1/y to get y->-infinity and then L'Hopitals.

  6. DLS
    • 3 years ago
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    |dw:1358405846397:dw| how will we solve this LHL anyway?

  7. hartnn
    • 3 years ago
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    where did 'a' come from ? did you try x=-1/y ?

  8. DLS
    • 3 years ago
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    its 0 not a nevermind sry :o and i dont want to use L hospital because i know answer is 4 :P

  9. hartnn
    • 3 years ago
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    why do i get -1 ? :P

  10. DLS
    • 3 years ago
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    LHL=-1 RHL=1 Limit does not exist :o

  11. hartnn
    • 3 years ago
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    confirmed with wolf ?

  12. DLS
    • 3 years ago
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    no with answer key..:o but how did we solve LHL to -1

  13. hartnn
    • 3 years ago
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    hmm....you can put h=-h in LHL to get same form as of RHL, that is of h->0+

  14. DLS
    • 3 years ago
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    |dw:1358406440539:dw| but this :/??

  15. hartnn
    • 3 years ago
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    you found RHL ? or not ? because above thing is just substituting x +h=a for that i suggested, x=-1/y instead.(and you get -1 then)

  16. DLS
    • 3 years ago
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    why x=-1/y? can u show clearly :/

  17. hartnn
    • 3 years ago
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    x=-1/y ---->-1/x = y to make the exponent of e as 'y' so that using L'Hopitals is easy (1+e^y)/(1-e^y) -----> e^y/(-e^y) ---->-1

  18. DLS
    • 3 years ago
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    itni si baat batane me aadha ghanta lagadia be :p

  19. hartnn
    • 3 years ago
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    :P but the limit doesn't exist, right ? you got that using LHL and RHL ?

  20. DLS
    • 3 years ago
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    yes..

  21. hartnn
    • 3 years ago
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    good :)

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