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Math Help - Limit

  1. #1
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    Limit

    If f is differentiable everywhere, then, in terms of f and a, what is the value of: lim x-->a (f(f(x))-f(f(a)))/(x-a)?
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  2. #2
    Senior Member ecMathGeek's Avatar
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    That equation simply finds the slope of a secant line between the points (x,f(x)) to (a,f(a)) as x approaches a. When x approaches a, the slope of the secant line begins to approach the slop of f(x) at a. In other words, it finds the slope at a or the derivative at a. Therefore, it equals f'(a).
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  3. #3
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    Quote Originally Posted by chaddy View Post
    If f is differentiable everywhere, then, in terms of f and a, what is the value of: lim x-->a (f(f(x))-f(f(a)))/(x-a)?
    Quote Originally Posted by ecMathGeek View Post
    That equation simply finds the slope of a secant line between the points (x,f(x)) to (a,f(a)) as x approaches a. When x approaches a, the slope of the secant line begins to approach the slop of f(x) at a. In other words, it finds the slope at a or the derivative at a. Therefore, it equals f'(a).
    I'm afraid not. Unless the f(f(x)) and f(f(a)) are typos, the answer is the derivative of f(f(x)) at x = a, NOT the derivative of f(x) at x = a ......

    Using the chain rule:

    [f(f(x))]' = f'(f(x)) \times f'(x). At x = a, this will equal f'(f(a)) \times f'(a). Making it a little neater, if you knew that f(a) = b, then it will equal f'(b) \times f'(a) ......
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  4. #4
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    Thanks for catching that.
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  5. #5
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    This question is a bit more complicated, I think.
    Because f is differentiable everywhere it is continuous everywhere.
    That means that \lim _{x \to a} \left( {f(x) - f(a)} \right) = 0.
    Now consider the following:
    \lim _{x \to a} \frac{{f\left( {f(x)} \right) - f\left( {f(a)} \right)}}<br />
{{x - a}} = \lim _{x \to a} \frac{{f\left( {f(x)} \right) - f\left( {f(a)} \right)}}<br />
{{f(x) - f(a)}}\lim _{x \to a} \frac{{f\left( x \right) - f\left( a \right)}}<br />
{{x - a}}
    \lim _{x \to a} \frac{{f\left( {f(x)} \right) - f\left( {f(a)} \right)}}<br />
{{x - a}} = f'\left( {f(a)} \right)f'(a)
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  6. #6
    Senior Member ecMathGeek's Avatar
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    I looked at it too quickly to notice the double f's. Thanks for catching that, Mr. Fantastic.
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