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Math Help - Compute the Limit: Question 2

  1. #1
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    Compute the Limit: Question 2

    I'm stuck on this one:

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

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

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

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

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

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

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

    = \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:

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

    u^4-1 = (u^2+1)(u^2-1)

    u^3-1 = (u-1)(u^2+u+1)

    \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; September 24th 2011 at 06:48 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 u^2+1 further, but you CAN factor u^2-1. try that, and see if something cancels...
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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 u^2+1 further, but you CAN factor u^2-1. try that, and see if something cancels...
    \text{Compute } \lim_{u \rightarrow 1 } \frac{u^4-1}{u^3-1}

    u^4-1 = (u^2+1)(u^2-1)= (u^2+1)(u+1)(u-1)

    u^3-1 = (u-1)(u^2+u+1)

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

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

    = \frac{1+1+1+1}{1+1+1}

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

    u^4-1 = (u^2+1)(u^2-1)= (u^2+1)(u+1)(u-1)

    u^3-1 = (u-1)(u^2+u+1)

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

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

    = \frac{1+1+1+1}{1+1+1}

    = \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:
    \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:
    \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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