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Math Help - Difference Quotient

  1. #1
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    Difference Quotient

    I have a problem in which i have to solve for a limit:

     lim  f(2+h) - f(2)/h
     h\rightarrow 0

     f(x) = 2 / x-1

    The divide sign is divided by the whole thing not just f(2) and that goes for f(x) also
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  2. #2
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    The limit is f'(2) you need to differeniate f(x) and plug in the appropriate value.

    Bobak
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  3. #3
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by snakeman11689 View Post
    I have a problem in which i have to solve for a limit:

     lim  f(2+h) - f(2)/h
     h\rightarrow 0

     f(x) = 2 / x-1

    The divide sign is divided by the whole thing not just f(2) and that goes for f(x) also
    f(x)=\frac{2}{x-1}?

    If so

    \lim_{\Delta{x}\to{0}}\frac{\frac{2}{x+\Delta{x}-1}-\frac{2}{x-1}}{\Delta{x}}

    Combining fractions we get

    \lim_{\Delta{x}\to{0}}\frac{\frac{-2\Delta{x}}{x^2+x(\Delta{x})-2x-\Delta{x}+1}}{\Delta{x}}=\lim_{\Delta{x}\to{0}}\fr  ac{-2}{x^2+x(\Delta{x})-2x-\Delta{x}+1}

    And direct substitution yields

    \frac{-2}{x^2-2x+1}=\frac{-2}{(x-1)^2}
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  4. #4
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    Quote Originally Posted by Mathstud28 View Post
    f(x)=\frac{2}{x-1}?

    If so

    \lim_{\Delta{x}\to{0}}\frac{\frac{2}{x+\Delta{x}-1}-\frac{2}{x-1}}{\Delta{x}}

    Combining fractions we get

    \lim_{\Delta{x}\to{0}}\frac{\frac{-2\Delta{x}}{x^2+x(\Delta{x})-2x-\Delta{x}+1}}{\Delta{x}}=\lim_{\Delta{x}\to{0}}\fr  ac{-2}{x^2+x(\Delta{x})-2x-\Delta{x}+1}

    And direct substitution yields

    \frac{-2}{x^2-2x+1}=\frac{-2}{(x-1)^2}
    hmm, the answer is -2 rite???
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  5. #5
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by snakeman11689 View Post
    hmm, the answer is -2 rite???
    \lim_{\Delta{x}\to{0}}\frac{f(x+\Delta{x})-f(x)}{\Delta{x}}=f'(x)\Rightarrow{\lim_{\Delta{x}\  to{0}}\frac{f(2+\Delta{x})-f(2)}{\Delta{x}}=f'(2)}

    So letting f(x)=\frac{2}{x-1}

    we have as shown above f'(x)=\frac{-2}{(x-1)^2}

    \therefore\lim_{\Delta{x}\to{0}}\frac{f(2+\Delta{x  })-f(2)}{\Delta{x}}=f'(2)=\frac{-2}{(2-1)^2}=-2
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  6. #6
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    Quote Originally Posted by Mathstud28 View Post
    \lim_{\Delta{x}\to{0}}\frac{f(x+\Delta{x})-f(x)}{\Delta{x}}=f'(x)\Rightarrow{\lim_{\Delta{x}\  to{0}}\frac{f(2+\Delta{x})-f(2)}{\Delta{x}}=f'(2)}

    So letting f(x)=\frac{2}{x-1}

    we have as shown above f'(x)=\frac{-2}{(x-1)^2}

    \therefore\lim_{\Delta{x}\to{0}}\frac{f(2+\Delta{x  })-f(2)}{\Delta{x}}=f'(2)=\frac{-2}{(2-1)^2}=-2

    got it , thank you so much for helping me
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