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Math Help - Having trouble with trig based limits

  1. #1
    TWN
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    Having trouble with trig based limits

    As h approaches 0,

    lim [cos((pi/2)+h) - cos(pi/2)] / h

    The answer is -1. Can someone please explain how to arrive at this answer?

    Are there special trig-based formulas I need to know to solve questions like this?
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    MHF Contributor Reckoner's Avatar
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    Re: Having trouble with trig based limits

    Quote Originally Posted by TWN View Post
    As h approaches 0,

    lim [cos((pi/2)+h) - cos(pi/2)] / h

    The answer is -1. Can someone please explain how to arrive at this answer?

    Are there special trig-based formulas I need to know to solve questions like this?
    Two special limits you should know are

    \lim_{x\to0}\frac{\sin x}{x}=1 and \lim_{x\to0}\frac{1-\cos x}{x}=0.

    In this case, we have

    \lim_{h\to0}\frac{\cos\left(\frac{\pi}2+h\right)-\cos\frac{\pi}2}h

    =\lim_{h\to0}\frac{\cos\frac\pi2\cos h-\sin\frac\pi2\sin h-0}h (sum and difference identity)

    =\lim_{h\to0}\frac{0\cdot\cos h-1\cdot\sin h}h

    =\lim_{h\to0}\frac{-\sin h}h

    =-\lim_{h\to0}\frac{\sin h}h

    =-1
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  3. #3
    TWN
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    Re: Having trouble with trig based limits

    Ah, I see. I was unaware of the sum and difference identities. Thank you!

    By the way, how do you write out the problems like that? It's so much easier to read than mine is!
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    Re: Having trouble with trig based limits

    Quote Originally Posted by TWN View Post
    Ah, I see. I was unaware of the sum and difference identities. Thank you!

    By the way, how do you write out the problems like that? It's so much easier to read than mine is!
    You can write in \LaTeX, just enclose your code between [tex] and [/tex] tags. For help with LaTeX, visit the LaTeX forum. There's a tutorial there.
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    Re: Having trouble with trig based limits

    Quote Originally Posted by TWN View Post
    As h approaches 0,

    lim [cos((pi/2)+h) - cos(pi/2)] / h

    The answer is -1. Can someone please explain how to arrive at this answer?

    Are there special trig-based formulas I need to know to solve questions like this?
    note the format of the limit ...

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

    ... this is the definition of f'(x)

    in this case, f(x) = \cos{x} and the limit is f'\left(\frac{\pi}{2}\right) = -\sin \left(\frac{\pi}{2}\right) = -1
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  6. #6
    TWN
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    Re: Having trouble with trig based limits

    That makes sense! Thank you both!
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