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Math Help - Calculus Limit Problem

  1. #1
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    Calculus Limit Problem

    I have a Calculus project. I have to find the following limit without using L'Hospital:

    Limit as x approaches a (x->a) of (cos(x)-cos(a))/(x-a)

    Thanks in advance.
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by djo201 View Post
    Thanks for the fast reply but I'm not getting it. Can you explain a little bit in more detail?

    Thanks

    Edit: I know the derivative of cos(x) or cos(a) is -sin(x) or -sin(a), respectively, but how do I get to that result?
    by defintion

    f'(a) = \lim_{x \to a} \frac {f(x) - f(a)}{x - a}

    here you have f(x) = \cos x

    so you can just say it is the derivative of cos(a) and just state that as the limit, namely, -sin(a)

    if you would rather calculate it manually, note that an alternate definition for the derivative is as follows

    f'(x) = \lim_{h \to 0} \frac {f(x + h) - f(x)}h

    it is easier to find this particular limit in that form

    getting your limit in that form is not as hard as you think

    let x = a + h, then \lim_{x \to a} \frac {\cos x - \cos a}{x - a} becomes \lim_{h \to 0} \frac { \cos (a + h) -  \cos a}h

    now proceed using the addition formula for cosine to simplify

    (note, you will need your special trig limits here)
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  3. #3
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    Quote Originally Posted by Jhevon View Post
    by defintion

    f'(a) = \lim_{x \to a} \frac {f(x) - f(a)}{x - a}

    here you have f(x) = \cos x

    so you can just say it is the derivative of cos(a) and just state that as the limit, namely, -sin(a)

    if you would rather calculate it manually, note that an alternate definition for the derivative is as follows

    f'(x) = \lim_{h \to 0} \frac {f(x + h) - f(x)}h

    it is easier to find this particular limit in that form

    getting your limit in that form is not as hard as you think

    let x = a + h, then \lim_{x \to a} \frac {\cos x - \cos a}{x - a} becomes \lim_{h \to 0} \frac { \cos (a + h) -  \cos a}h

    now proceed using the addition formula for cosine to simplify

    (note, you will need your special trig limits here)
    Thank you very much. Now I get it.
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    Ok, one last question, why does x->a become h->0?
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by djo201 View Post
    Ok, one last question, why does x->a become h->0?
    x = a + h

    \Rightarrow h = x - a

    now as x gets close to a, what does h get close to?
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  6. #6
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    Quote Originally Posted by Jhevon View Post
    x = a + h

    \Rightarrow h = x - a

    now as x gets close to a, what does h get close to?
    Thanks again.
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