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Thread: Difference Quotient

  1. #1
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    Difference Quotient

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

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

    $\displaystyle 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 $\displaystyle f'(2)$ you need to differeniate $\displaystyle f(x)$ 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:

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

    $\displaystyle 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
    $\displaystyle f(x)=\frac{2}{x-1}$?

    If so

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

    Combining fractions we get

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

    And direct substitution yields

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

    If so

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

    Combining fractions we get

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

    And direct substitution yields

    $\displaystyle \frac{-2}{x^2-2x+1}=\frac{-2}{(x-1)^2}$
    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???
    $\displaystyle \lim_{\Delta{x}\to{0}}\frac{f(x+\Delta{x})-f(x)}{\Delta{x}}=f'(x)\Rightarrow{\lim_{\Delta{x}\ to{0}}\frac{f(2+\Delta{x})-f(2)}{\Delta{x}}=f'(2)}$

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

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

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

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

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

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

    got it , thank you so much for helping me
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