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Math Help - Limits Question

  1. #1
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    Limits Question

    If I have this problem:

    lim as h -> 0
    <br />
\frac{sin( \frac{\pi}{2} + h) - 1}{h}<br />

    And I have these answer choices:

    A. f'(\frac{\pi}{2}), f(x)=sinx
    B. f'(\frac{\pi}{2}), f(x)=\frac{sinx}{x}
    C. f'(1), f(x)=sinx
    D. f'(1), f(x)=sin(x + \frac{\pi}{2})
    E. f'(0), f(x)=sinx

    What does this really mean? like, I don't understand what f'(0) would imply, for instance.
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  2. #2
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    Quote Originally Posted by Lord Darkin View Post
    If I have this problem:

    lim as h -> 0
    <br />
\frac{sin( \frac{\pi}{2} + h) - 1}{h}<br />

    And I have these answer choices:

    A. f'(\frac{\pi}{2}), f(x)=sinx
    B. f'(\frac{\pi}{2}), f(x)=\frac{sinx}{x}
    C. f'(1), f(x)=sinx
    D. f'(1), f(x)=sin(x + \frac{\pi}{2})
    E. f'(0), f(x)=sinx

    What does this really mean? like, I don't understand what f'(0) would imply, for instance.


    If a function f(x) is defined on a point x_0 and in some neighborhood of this point , the definition of the derivative of f in this point is:

    f'(x_0):=\lim_{h\to 0}\frac{f(x_0+h)-f(x_0)}{h} , in case this limits exists and is finite.

    Can you see now what option to choose?

    Tonio
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  3. #3
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    So ... f(x_{0}) = 1

    Would it be choice a? Since if I put in pi/2, then that means that f(x) would equal 1 since sin(90)=1.
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  4. #4
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    Yes, that is correct.

    (Although I would prefer " sin(\pi/2)= 1" to "sin(90)= 1"! You need to start thinking in radians, not degrees.)
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  5. #5
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    Yeah, I would ahve thought that. Thanks.
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