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Math Help - Harder limits

  1. #1
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    Harder limits

    The function f is differentiable at a. Find \lim_{h\to 0}\frac{f(a+ph)-f(a-ph)}{h}

    I'd assume its something to do with manipulating the definition \lim_{h\to 0}\frac{f(a+h)-f(a)}{h} but have no idea how.
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  2. #2
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    Quote Originally Posted by vuze88 View Post
    The function f is differentiable at a. Find \lim_{h\to 0}\frac{f(a+ph)-f(a-ph)}{h}

    I'd assume its something to do with manipulating the definition \lim_{h\to 0}\frac{f(a+h)-f(a)}{h} but have no idea how.
    \lim_{h\to 0}\frac{f(a+ph)-f(a-ph)}{h}=p\lim_{h\to 0}\frac{f(a+ph)-f(a-ph)+f(a)-f(a)}{ph}\\

    could you finish it?
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  3. #3
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    so is the answer 2pf'(a)
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  4. #4
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    Quote Originally Posted by felper View Post
    \lim_{h\to 0}\frac{f(a+ph)-f(a-ph)}{h}=p\lim_{h\to 0}\frac{f(a+ph)-f(a-ph)+f(a)-f(a)}{ph}\\

    could you finish it?
    p\lim_{h\to 0}\frac{f(a+ph)-f(a-ph)+f(a)-f(a)}{ph}
    =p\lim_{h\to 0}\frac{f(a+ph)-f(a)}{ph}+\frac{f(a)-f(a-ph)}{ph}
    =p\lim_{h\to 0}\frac{f(a+ph)-f(a)}{ph}+\frac{f(a-ph)-f(a)}{-ph}=2pf'(a)
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