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Math Help - Difficult Limit

  1. #1
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    Difficult Limit

    Limit as x goes towards 0 for the function (e^x-1)^2/(sin(e^x-1))

    Any help would be much appreciated
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  2. #2
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    Re: Difficult Limit

    Quote Originally Posted by DylanRenke View Post
    Limit as x goes towards 0 for the function (e^x-1)^2/(sin(e^x-1))

    Any help would be much appreciated
    To simplify things let

    u=e^x-1

    If we take the limit as x \to 0 this is the same as the limit as u \to 0

    Now this gives you the limit

    \lim_{x \to 0}\frac{(e^{x}-1)^2}{\sin(e^x-1)}=\lim_{u \to 0}\frac{u^2}{\sin(u)}

    This gives

    \lim_{u \to 0}\frac{u}{\sin(u)}\cdot u=1 \cdot 0 =0
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  3. #3
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    Re: Difficult Limit

    Why doesnt that last line just give you 0/0? because when u approaches 0, it equals 0, and when sin(u) approaches 0, its zero as well?
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    Re: Difficult Limit

    Quote Originally Posted by DylanRenke View Post
    Why doesnt that last line just give you 0/0? because when u approaches 0, it equals 0, and when sin(u) approaches 0, its zero as well?
    This is a well known limit and its value should be known. The proof is a bit lengthy and requires a bit of trigonometry and the use of the squeeze theorem.

    Do you need a proof?
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  5. #5
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    Re: Difficult Limit

    It would be quiet helpful, because i really do not understand it.
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  6. #6
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    Re: Difficult Limit

    Actually, i have found that the professor told us to memorize that limit. Thank you for the help
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