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Math Help - Another Derivative Problem

  1. #1
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    Another Derivative Problem

    Let f be differentiable at a and let {x_n} and {z_n} be two sequences converging to a such that x_n < a < z_n for n \in \mathbb{N}. Prove that
    \lim_{n \to \infty} \frac{f(x_n)-f(z_n)}{x_n-z_n} = f^{'}(a).
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  2. #2
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    This looks like a result of the Mean Value Theorem...
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  3. #3
    Senior Member Sambit's Avatar
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    x_n and z_n converges to a and x_n<a<z_n. This gives lim_{n \to \infty} x_n = a = lim_{n \to \infty} z_n. In other words, lim_{n \to \infty}(x_n-z_n)=0. Again, lim_{n \to \infty} x_n = lim_{n \to \infty} z_n; consequently, lim_{n \to \infty} f(x_n) - lim_{n \to \infty} f(z_n) or lim_{n \to \infty} (f(x_n) - f(z_n) ) converges to 0.
    Thus the expression can be thought of the form \lim_{x \to a} \frac{f(x)-f(a)}{x-a} that is f'(a).
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  4. #4
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    Hey,

    The question doesn't allow us to use mean value theorem directly to the function.

    Anyway,the hint given to this question is that using \frac{f(x_n)-f(z_n)}{x_n-z_n} = \frac{f(x_n)-f(a)}{x_n-a} \frac{x_n-a}{x_n-z_n} + \frac{f(z_n)-f(a)}{z_n-a} \frac{a-z_n}{x_n-z_n}, show that this lie between \frac{f(x_n)-f(a)}{x_n-a} and \frac{f(z_n)-f(a)}{z_n-a}.

    I've tried using this hint, but I can't get anywhere between \frac{f(x_n)-f(a)}{x_n-a} and \frac{f(z_n)-f(a)}{z_n-a}. And if I'm not wrong, we can't assume that the function is monotonic as there might be cases where f(x_n) \leq f(z_n) for some n \in \mathbb{N} and f(x_n) \geq f(z_n) for the remaining integers in \mathbb{N}.

    So any idea on how to continue from the hint above?

    Thanks in advance.
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  5. #5
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    May be you can use this

    <br />
f'_{max}(a)=max( \; \frac{f(x_n)-f(a)}{x_n-a}\; ; \; \frac{f(z_n)-f(a)}{z_n-a} \; )<br />

    <br />
f'_{min}(a)=min( \; \frac{f(x_n)-f(a)}{x_n-a}\; ; \; \frac{f(z_n)-f(a)}{z_n-a} \; )<br />
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  6. #6
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    \frac{f(x_n)-f(z_n)}{x_n-z_n}
    = \frac{f(x_n)-f(a)}{x_n-a} \frac{x_n-a}{x_n-z_n} + \frac{f(z_n)-f(a)}{z_n-a} \frac{a-z_n}{x_n-z_n}
    \leq max(\frac{f(x_n)-f(a)}{x_n-a} , \frac{f(z_n)-f(a)}{z_n-a}) \frac{x_n-a}{x_n-z_n} + max(\frac{f(x_n)-f(a)}{x_n-a} , \frac{f(z_n)-f(a)}{z_n-a}) \frac{a-z_n}{x_n-z_n}
    = max(\frac{f(x_n)-f(a)}{x_n-a} , \frac{f(z_n)-f(a)}{z_n-a}).
    Similarly,
    \frac{f(x_n)-f(z_n)}{x_n-z_n} \geq min(\frac{f(x_n)-f(a)}{x_n-a} , \frac{f(z_n)-f(a)}{z_n-a}).
    Then min(\frac{f(x_n)-f(a)}{x_n-a} , \frac{f(z_n)-f(a)}{z_n-a}) \leq \frac{f(x_n)-f(z_n)}{x_n-z_n} \leq max(\frac{f(x_n)-f(a)}{x_n-a} , \frac{f(z_n)-f(a)}{z_n-a}).
    Since lim_{n \to \infty} x_n = a and lim_{n \to \infty} z_n = a, lim_{n \to \infty} \frac{f(x_n)-f(a)}{x_n-a} = f'(a) and lim_{n \to \infty} \frac{f(z_n)-f(a)}{z_n-a} = f'(a). Then lim_{n \to \infty} max(\frac{f(x_n)-f(a)}{x_n-a} , \frac{f(z_n)-f(a)}{z_n-a}) =  f'(a) and lim_{n \to \infty} max(\frac{f(x_n)-f(a)}{x_n-a} , \frac{f(z_n)-f(a)}{z_n-a}) =  f'(a). Then by squeeze theorem, lim_{n \to \infty} \frac{f(x_n)-f(z_n)}{x_n-z_n} = f'(a).

    I hope that this is correct. Any opinion?
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  7. #7
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    Or you could say that

    <br />
f'_{max}(a)= \; \frac{f(z_n)-f(a)}{z_n-a} <br />

    and then

    <br />
f'_{min}(a)= \; \frac{f(x_n)-f(a)}{x_n-a} .<br />

    Or vice versa.
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