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anonymous

  • 5 years ago

Lim(x->0)[1/x]^sinx

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  1. anonymous
    • 5 years ago
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    that form is \[\infty\]^0, so u have to use exp function

  2. anonymous
    • 5 years ago
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    The answer given is 1

  3. anonymous
    • 5 years ago
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    Why don't you do a series expansion in x and then send the function to zero?

  4. anonymous
    • 5 years ago
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    Actually, I should have worked it out before opening my mouth.

  5. anonymous
    • 5 years ago
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    exp[lim(x->0) sin x . 1/x] exp[lim(x->0) (ln x)/sin x] and then do l'hospital

  6. anonymous
    • 5 years ago
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    Actually, what I said will work, but you'd have to prove other limits first.

  7. anonymous
    • 5 years ago
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    i do thinking about the concept before typing

  8. anonymous
    • 5 years ago
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    What are you thinking, iamignorant?

  9. anonymous
    • 5 years ago
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    Yes, I have got the idea that suzi gave. With that idea I am coming at e^(0x1)

  10. anonymous
    • 5 years ago
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    So that is the answer that I needed. Actually I did everything he did except applying lHospital

  11. anonymous
    • 5 years ago
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    ok

  12. anonymous
    • 5 years ago
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    Thank you all

  13. anonymous
    • 5 years ago
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    np

  14. anonymous
    • 5 years ago
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    is there a rank higher than HERO? =D like a grandmaster or something

  15. anonymous
    • 5 years ago
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    Who knows? This site's operations are a mystery half the time.

  16. anonymous
    • 5 years ago
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    i think GOD

  17. anonymous
    • 5 years ago
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    =D LOL

  18. anonymous
    • 5 years ago
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    It's dead today...

  19. anonymous
    • 5 years ago
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    Incidentally, I set the RHS equal to y then took the log of both sides, put the result into an indeterminate form, used L'Hopital's rule and took the limit to zero after I suggested expanding the function. Just sayin'...

  20. anonymous
    • 5 years ago
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    What's going on BMF?

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spraguer (Moderator)
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