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Thread: Another Hard Limit Question

  1. #1
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    Another Hard Limit Question

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

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

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