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Math Help - special limits of trig functions

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    special limits of trig functions

    compute using special limits of trig functions: lim x->pie (cosx+1)/(x-pie)
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by myoplex11 View Post
    compute using special limits of trig functions: lim x->pie (cosx+1)/(x-pie)
    First of all, it's "pi" not "pie."

    I'm not sure how to break this into anything resembling a "special limit" but it's not hard to do directly:

    \lim_{x \to \pi} \frac{cosx + 1}{x - \pi}

    Note that this is of the form 0/0, so use L'Hopital's rule:

    \lim_{x \to \pi} \frac{cosx + 1}{x - \pi} = \lim_{x \to \pi} \frac{-sinx}{1} = 0

    -Dan
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    Quote Originally Posted by topsquark View Post
    First of all, it's "pi" not "pie."

    I'm not sure how to break this into anything resembling a "special limit" but it's not hard to do directly:

    \lim_{x \to \pi} \frac{cosx + 1}{x - \pi}

    Note that this is of the form 0/0, so use L'Hopital's rule:

    \lim_{x \to \pi} \frac{cosx + 1}{x - \pi} = \lim_{x \to \pi} \frac{-sinx}{1} = 0

    -Dan
    You can do this,
    \lim_{x\to \pi} \frac{1-\cos (x- \pi)}{x-\pi}
    Using the identity,
    \cos x=-\cos (x-\pi)
    Now this is a composition function,
    f(x)=\frac{1-\cos x}{x}
    g(x)=x-\pi
    And,
    \lim_{x\to \pi}g(x)=0
    Thus the limit of,
    \lim_{x\to \pi}\frac{1-\cos (x-\pi)}{x-\pi}=\lim_{x\to 0}\frac{1-\cos x}{x}=0
    By composition rule for limits.
    That is a standard limit.
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by ThePerfectHacker View Post
    You can do this,
    \lim_{x\to \pi} \frac{1-\cos (x- \pi)}{x-\pi}
    Using the identity,
    \cos x=-\cos (x-\pi)
    Now this is a composition function,
    f(x)=\frac{1-\cos x}{x}
    g(x)=x-\pi
    And,
    \lim_{x\to \pi}g(x)=0
    Thus the limit of,
    \lim_{x\to \pi}\frac{1-\cos (x-\pi)}{x-\pi}=\lim_{x\to 0}\frac{1-\cos x}{x}=0
    By composition rule for limits.
    That is a standard limit.
    Clever! I like it.

    -Dan
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