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Math Help - How to find this limit?

  1. #1
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    How to find this limit?

    lim (cube rt (8+h) - 2) / h
    h->0
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  2. #2
    Ted
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    Think about the definition of the derivative ..
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  3. #3
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    It's supposed to be solved without using L'Hopital.
    Maybe using rationalization, I don't know...
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  4. #4
    Ted
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    Quote Originally Posted by sodk View Post
    It's supposed to be solved without using L'Hopital.
    Maybe using rationalization, I don't know...
    Did you see post #2 ?
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  5. #5
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    Or just use good old L`l˘pital rule

    \lim_{h\to 0} \frac{{\sqrt[3]{{8 + h}} - 2}}{h} \Rightarrow \lim_{h \to 0} \frac{{\frac{1}{{3\sqrt[3]{{\left( {8 + h} \right)^2 }}}}}}{1} \Rightarrow \frac{1}{{3\sqrt[3]{{\left( {8 + 0} \right)^2 }}}} \Rightarrow \frac{1}{{3\sqrt[3]{{64}}}} \Rightarrow \frac{{\sqrt[3]{{64^2 }}}}{{192}} \Rightarrow \frac{{\sqrt[3]{{4096}}}}{{192}} \Rightarrow \frac{{16}}{{192}}  \Rightarrow \underline{\underline {{\rm{ }}\frac{1}{{12}}{\rm{ }}}}

    Mmm Latex is strange ^^
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  6. #6
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    Quote Originally Posted by sodk View Post
    lim (cube rt (8+h) - 2) / h
    h->0

    expanding on Ted's excellent hint ...

     <br />
f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h}<br />
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  7. #7
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    Hello, sodk!

    Here is Ted's suggestion . . . in baby-steps.

    Recall that: . (a-b)(a^2 + ab + b^2)\;=\;a^3-b^3


    Find: . \lim_{h\to0} \frac{\sqrt[3]{8+h} - 2}{h}

    Multiply by: . \frac{\sqrt[3]{(8+h)^2} + 2\sqrt[3]{8+h} + 4} {\sqrt[3]{(8+h)^2} + 2\sqrt[3]{8+h} + 4}

    \frac{\sqrt[3]{8+h} - 2}{h} \cdot\frac{\sqrt[3]{(8+h)^2} + 2\sqrt[3]{8+h} + 4} {\sqrt[3]{(8+h)^2} + 2\sqrt[3]{8+h} + 4} \;\;=\;\;\frac{(8+h) - 8}<br />
{h\left[\sqrt[3]{(8+h)^2} + 2\sqrt[3]{8+h} + 4\right]}

    . . . =\;\;\frac{h} {h\left[\sqrt[3]{(8+h)^2} + 2\sqrt[3]{8+h} + 4\right]} \;\;=\;\;\frac{1}{\sqrt[3]{(8+h)^2} + 2\sqrt[3]{8+h} + 4}


    Therefore: . \lim_{h\to0}\left[ \frac{1}{\sqrt[3]{(8+h)^2} + 2\sqrt[3]{8+h} + 4}\right]\;\;=\;\;\frac{1}{(\sqrt[3]{8})^2 + 2\sqrt[3]{8} + 4}\;\;=\;\;\frac{1}{4+4+4}\;\;=\;\;\frac{1}{12}

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  8. #8
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    match up ...

     <br />
f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h}<br />

    to

    f'(8) = \lim_{h \to 0} \frac{\sqrt[3]{8+h} - \sqrt[3]{8}}{h}<br />

    and realize that f(x) = \sqrt[3]{x}

    f'(x) = \frac{1}{3\sqrt[3]{x^2}}

    and the desired limit is the value of f'(8) ...

    f'(8) = \frac{1}{12}
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  9. #9
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    Quote Originally Posted by Soroban View Post
    Hello, sodk!

    Here is Ted's suggestion . . . in baby-steps.

    Recall that: . (a-b)(a^2 + ab + b^2)\;=\;a^3-b^3


    Multiply by: . \frac{\sqrt[3]{(8+h)^2} + 2\sqrt[3]{8+h} + 4} {\sqrt[3]{(8+h)^2} + 2\sqrt[3]{8+h} + 4}

    \frac{\sqrt[3]{8+h} - 2}{h} \cdot\frac{\sqrt[3]{(8+h)^2} + 2\sqrt[3]{8+h} + 4} {\sqrt[3]{(8+h)^2} + 2\sqrt[3]{8+h} + 4} \;\;=\;\;\frac{(8+h) - 8} {h\left[\sqrt[3]{(8+h)^2} + 2\sqrt[3]{8+h} + 4\right]}" alt="
    {h\left[\sqrt[3]{(8+h)^2} + 2\sqrt[3]{8+h} + 4\right]}" />

    . . . =\;\;\frac{h} {h\left[\sqrt[3]{(8+h)^2} + 2\sqrt[3]{8+h} + 4\right]} \;\;=\;\;\frac{1}{\sqrt[3]{(8+h)^2} + 2\sqrt[3]{8+h} + 4}


    Therefore: . \lim_{h\to0}\left[ \frac{1}{\sqrt[3]{(8+h)^2} + 2\sqrt[3]{8+h} + 4}\right]\;\;=\;\;\frac{1}{(\sqrt[3]{8})^2 + 2\sqrt[3]{8} + 4}\;\;=\;\;\frac{1}{4+4+4}\;\;=\;\;\frac{1}{12}
    That was exactly what I wanted! Unfortunately your post came late, I've solved it this exact way. My mistake was just trying to rationalize the numerator incorrectly. The trick is the factors (a-b)(a^2 + ab + b^2)\;=\;a^3-b^3
    Anyway thanks.
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