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Math Help - Limit Question

  1. #1
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    Limit Question

    I'm confused:

    Why is the limit(as x -> 0) sin(x)/(x-sin(x)) = infinity?

    I have done this 10 times now, and I get that it equals -1.

    We have an indeterminant form when we plug 0 in (0/0). Thus, I can use l'hopitals rule.

    Then:

    cos(x)/(1-cos(x)) = 1/0...doesn't work, so do l'hopitals again.

    -sin(x)/sin(x) = 0/0; use l'hopitals again

    -cos(x)/cos(x) = -1/1 = -1

    So, how does it equal infinity?
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  2. #2
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    Quote Originally Posted by Ideasman View Post
    I'm confused:

    Why is the limit(as x -> 0) sin(x)/(x-sin(x)) = infinity?

    I have done this 10 times now, and I get that it equals -1.

    We have an indeterminant form when we plug 0 in (0/0). Thus, I can use l'hopitals rule.

    Then:

    cos(x)/(1-cos(x)) = 1/0...doesn't work, so do l'hopitals again.

    -sin(x)/sin(x) = 0/0; use l'hopitals again

    -cos(x)/cos(x) = -1/1 = -1

    So, how does it equal infinity?
    Do not use L'Hopitals rule,
    You have,
    \frac{\sin x}{x-\sin x}
    Divide by \sin x
    Thus,
    \frac{1}{\frac{x}{\sin x}-1}
    So basically, the limit,
    \lim_{x\to 0}\frac{x}{\sin x}=1
    One of standard limits.
    Thus, you have division by zero with non-zero numerator thus the limit is infinite. (does not exist).
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  3. #3
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    Why can't you use l'hopitals rule? It is an indeterminant form, 0/0.
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  4. #4
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    Quote Originally Posted by Ideasman View Post
    Why can't you use l'hopitals rule? It is an indeterminant form, 0/0.
    You can, but you used it a second time when the form was 1/0, which was incorrect. Arriving at 1/0 actually would imply the answer is infinity.
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  5. #5
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    Quote Originally Posted by Ideasman View Post
    Why can't you use l'hopitals rule? It is an indeterminant form, 0/0.
    You can, but I rather not.

    You mistake is that after you use the L-Hopital rule once you have 1/0 which is limits of infinite. YOU CANNOT USE RULE HERE. Its proof is based on the fact that both are zero!
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