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Thread: Compute the Limit: Question 2

  1. #1
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    Compute the Limit: Question 2

    I'm stuck on this one:

    $\displaystyle \text{Compute } \lim_{u \rightarrow 1 } \frac{u^4-1}{u^3-1}$

    $\displaystyle = \frac{1^4-1}{1^3-1}$

    $\displaystyle = \frac{0}{0 } \text{Indeterminate}$

    $\displaystyle \lim_{u \rightarrow 1 } \frac{u^4-1}{u^3-1}$

    $\displaystyle = \lim_{u \rightarrow 1 } \frac{(u-1)(u+1)(u-1)(u-1)}{(u-1)(u+1)(u+1)}$

    $\displaystyle = \lim_{u \rightarrow 1 } \frac{(u-1)(u-1)}{(u+1)}$

    $\displaystyle = \frac{(1-1)(1-1)}{(1+1) }$

    $\displaystyle = \frac{0}{2}$
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  2. #2
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    Re: Compute the Limit: Question 2

    Not good, Sparky. Please factor numerator and denominator again. Both are incorrect.

    Not good, Sparky. Why did you substitute the limit value into a function you do NOT know is continuous? Never do that.

    Do a little less manipulating little symbols and a lot more thinking.
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  3. #3
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    Re: Compute the Limit: Question 2

    Ok, here is another attempt at the following question:

    $\displaystyle \text{Compute } \lim_{u \rightarrow 1 } \frac{u^4-1}{u^3-1}$

    $\displaystyle u^4-1 = (u^2+1)(u^2-1)$

    $\displaystyle u^3-1 = (u-1)(u^2+u+1)$

    $\displaystyle \lim_{u \rightarrow 1 } \frac{(u^2+1)(u^2-1)}{(u-1)(u^2+u+1)}$

    Where did I go wrong with my factorization?
    Last edited by sparky; Sep 24th 2011 at 05:48 PM.
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  4. #4
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    Re: Compute the Limit: Question 2

    so far, so good. you won't be able to factor $\displaystyle u^2+1$ further, but you CAN factor $\displaystyle u^2-1$. try that, and see if something cancels...
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  5. #5
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    Re: Compute the Limit: Question 2

    Quote Originally Posted by Deveno View Post
    so far, so good. you won't be able to factor $\displaystyle u^2+1$ further, but you CAN factor $\displaystyle u^2-1$. try that, and see if something cancels...
    $\displaystyle \text{Compute } \lim_{u \rightarrow 1 } \frac{u^4-1}{u^3-1}$

    $\displaystyle u^4-1 = (u^2+1)(u^2-1)= (u^2+1)(u+1)(u-1)$

    $\displaystyle u^3-1 = (u-1)(u^2+u+1)$

    $\displaystyle = \lim_{u \rightarrow 1 } \frac{(u^2+1)(u+1)(u-1)}{(u-1)(u^2+u+1)}$

    $\displaystyle = \lim_{u \rightarrow 1 } \frac{(u^2+1)(u+1)}{(u^2+u+1)}$

    $\displaystyle = \frac{1+1+1+1}{1+1+1}$

    $\displaystyle = \frac{4}{3}$

    Is this correct?
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  6. #6
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    Re: Compute the Limit: Question 2

    It is a beautiful thing.
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  7. #7
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    Re: Compute the Limit: Question 2

    Quote Originally Posted by sparky View Post
    $\displaystyle \text{Compute } \lim_{u \rightarrow 1 } \frac{u^4-1}{u^3-1}$

    $\displaystyle u^4-1 = (u^2+1)(u^2-1)= (u^2+1)(u+1)(u-1)$

    $\displaystyle u^3-1 = (u-1)(u^2+u+1)$

    $\displaystyle = \lim_{u \rightarrow 1 } \frac{(u^2+1)(u+1)(u-1)}{(u-1)(u^2+u+1)}$

    $\displaystyle = \lim_{u \rightarrow 1 } \frac{(u^2+1)(u+1)}{(u^2+u+1)}$

    $\displaystyle = \frac{1+1+1+1}{1+1+1}$

    $\displaystyle = \frac{4}{3}$

    Is this correct?
    your answer is correct, but the work is not. it is lucky for you that (1+1)(1+1) = 1+1+1+1.
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  8. #8
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    Re: Compute the Limit: Question 2

    Quote Originally Posted by Deveno View Post
    your answer is correct, but the work is not. it is lucky for you that (1+1)(1+1) = 1+1+1+1.
    My work is not correct? Where did I go wrong?
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  9. #9
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    Re: Compute the Limit: Question 2

    In this step you're substituting wrong, you wrote:
    $\displaystyle \lim_{u\to 1} \frac{(u^2+1)(u+1)}{u^2+u+1}=\frac{1+1+1+1}{1+1+1} =\frac{4}{3}$
    It has to be:
    $\displaystyle \lim_{u\to 1} \frac{(u^2+1)(u+1)}{u^2+u+1}=\frac{(1+1)(1+1)}{1+1 +1}=\frac{4}{3}$

    Do you notice your mistake? Offcourse like Deveno said it's lucky that (1+1)(1+1)=1+1+1+1, but in other cases ...
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