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Math Help - Another Hard Limit Question

  1. #1
    Senior Member polymerase's Avatar
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    Another Hard Limit Question

    Determine the values of a and b if \lim_{x\to{0}}\frac{(ax+b)^{\frac{1}{3}}-2}{x} = \frac{5}{12}
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  2. #2
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    Quote Originally Posted by polymerase View Post
    Determine the values of a and b if \lim_{x\to{0}}\frac{(ax+b)^{\frac{1}{3}}-2}{x} = \frac{5}{12}
    We require that \lim_{x\to 0}(ax+b)^{1/3} - 2 = 0 because otherwise the limit would not exist. But that means now b^{1/3} = 2 \implies b = 8.
    So we need to solve,
    \lim_{x\to 0}\frac{(ax+8)^{1/3} - 2}{x} = 0.

    To do this we use the fact that f(x)=(ax+8)^{1/3} is differenciable at x=0 and so f(x) = f(0)+xf'(0)+E(x) where E(x)/x\to 0. Thus, f(x) = 2 + ax/12 + E(x).

    Thus,
    \lim_{x\to 0}\frac{(ax+8)^{1/3} - 2}{x} = \lim_{x\to 0}\frac{2+ax/12 + E(x) - 2}{x} = \frac{a}{12}.
    So a=5
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