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Math Help - Different kind of Limit Problem

  1. #1
    Senior Member polymerase's Avatar
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    Different kind of Limit Problem

    Suppost that f is a differentiable function such that f'(8)=2. Without using L'Hospital's Rule, find the value of \displaystyle\lim_{x\to 8}\frac{f(x)-f(8)}{x^{\frac{1}{3}}-2}
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by polymerase View Post
    Suppost that f is a differentiable function such that f'(8)=2. Without using L'Hospital's Rule, find the value of \displaystyle\lim_{x\to 8}\frac{f(x)-f(8)}{x^{\frac{1}{3}}-2}
    does this help?


    f'(8) = 2 = \lim_{x \to 8} \frac {f(x) - f(8)}{x - 8} = \lim_{x \to 8} \frac {f(x) - f(8)}{\left( x^{\frac 13} - 2 \right) \left( x^{\frac 23} + 2x^{\frac 13} + 4 \right) }
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  3. #3
    Senior Member polymerase's Avatar
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    Quote Originally Posted by Jhevon View Post
    does this help?


    f'(8) = 2 = \lim_{x \to 8} \frac {f(x) - f(8)}{x - 8} = \lim_{x \to 8} \frac {f(x) - f(8)}{\left( x^{\frac 13} - 2 \right) \left( x^{\frac 23} + 2x^{\frac 13} + 4 \right) }
    No not really, I still don't understand how to solve it. I get how you factored....thats about it
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by polymerase View Post
    No not really, I still don't understand how to solve it. I get how you factored....thats about it
    do you not see the limit we want in there?

    2 =  \lim_{x \to 8} \frac {f(x) - f(8)}{\left( x^{\frac 13} - 2 \right) \left( x^{\frac 23} + 2x^{\frac 13} + 4 \right) } = \lim_{x \to 8} \frac {f(x) - f(8)}{x^{\frac 13} - 2} \cdot \frac 1{x^{\frac 23} + 2x^{\frac 13} + 4}

    what about now?
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  5. #5
    Senior Member polymerase's Avatar
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    Quote Originally Posted by Jhevon View Post
    do you not see the limit we want in there?

    2 = \lim_{x \to 8} \frac {f(x) - f(8)}{\left( x^{\frac 13} - 2 \right) \left( x^{\frac 23} + 2x^{\frac 13} + 4 \right) } = \lim_{x \to 8} \frac {f(x) - f(8)}{x^{\frac 13} - 2} \cdot \frac 1{x^{\frac 23} + 2x^{\frac 13} + 4}

    what about now?
    Ahhh i get it...so the answer is 24 right? Ok back track a little...how did you know that 2 = the limit?
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by polymerase View Post
    Ahhh i get it...so the answer is 24 right? Ok back track a little...how did you know that 2 = the limit?
    yes, it's 24.

    we were told f'(8) = 2. i used the limit definition of the derivative
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  7. #7
    Senior Member polymerase's Avatar
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    Quote Originally Posted by Jhevon View Post
    yes, it's 24.

    we were told f'(8) = 2. i used the limit definition of the derivative
    Ahhhhh....i get it completely now. Thanks
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