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Math Help - Evaluating Limits with rationals.

  1. #1
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    Evaluating Limits with rationals.

    I got a question on my test today and I spent a great deal of time on the question.
    lim as x -> 27 (27-x)/[(x^1/3)-3]

    I was able to approximate the limit by evaluating left and right of 27; however, there was another method where a change of variable was used, for example let u=(x^1/3).

    I was unsure how to proceed this way. any help would be appreciated.
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  2. #2
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    Quote Originally Posted by darksoulzero View Post
    I got a question on my test today and I spent a great deal of time on the question.
    lim as x -> 27 (27-x)/[(x^1/3)-3]

    I was able to approximate the limit by evaluating left and right of 27; however, there was another method where a change of variable was used, for example let u=(x^1/3).

    I was unsure how to proceed this way. any help would be appreciated.
    Let \displaystyle {u = x^{1/3} \Rightarrow x = u^3}. Then your limit becomes \displaystyle{\lim_{u \to 3} \frac{27 - u^3}{u - 3}} which should be simple for you to do.
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  3. #3
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    More help:

    (a^3 - b^3)=(a-b)(a^2+ab+b^2)
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    Quote Originally Posted by darksoulzero View Post
    I got a question on my test today and I spent a great deal of time on the question.
    lim as x -> 27 (27-x)/[(x^1/3)-3]

    I was able to approximate the limit by evaluating left and right of 27; however, there was another method where a change of variable was used, for example let u=(x^1/3).

    I was unsure how to proceed this way. any help would be appreciated.
    You can factor out 3-x^{1/3} in the numerator and then cancel that factor against the denominator like this:

    \lim\limits_{x\to 27}\frac{27-x}{x^{1/3}-3}=\lim\limits_{x\to 27}\frac{(3-x^{1/3})(3^2+3x^{1/3}+x^{2/3})}{x^{1/3}-3}=\lim\limits_{x\to 27}\left[-(3^2+3x^{1/3}+x^{2/3})\right]=-27
    Last edited by Failure; July 10th 2010 at 12:16 AM. Reason: removed factor 27 to fit original problem statement
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    Quote Originally Posted by Failure View Post
    You can factor out 3-x^{1/3} in the numerator and then cancel that factor against the denominator like this:

    \lim\limits_{x\to 27}\frac{27(27-x)}{x^{1/3}-3}=\lim\limits_{x\to 27}\frac{27(3-x^{1/3})(3^2+3x^{1/3}+x^{2/3})}{x^{1/3}-3}=\lim\limits_{x\to 27}\left[-27(3^2+3x^{1/3}+x^{2/3})\right]=-729
    I think you'll find that there's a mistake somewhere here because -729 is not the answer.
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  6. #6
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    Quote Originally Posted by mr fantastic View Post
    I think you'll find that there's a mistake somewhere here because -729 is not the answer.
    yeah, the answer is actually -27, but thanks for assistance gentlemen I did not realize that you could have factored out the -27.
    Last edited by mr fantastic; July 10th 2010 at 02:59 AM. Reason: The new question tagged to the thread was moved to a new thread.
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  7. #7
    Super Member Failure's Avatar
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    Quote Originally Posted by mr fantastic View Post
    I think you'll find that there's a mistake somewhere here because -729 is not the answer.
    Thank you for the hint: I had mistakenly added the factor 27 . I will fix that mistake in my reply to fit the original problem statement (although, what I had written ist not, strictly speaking, wrong: it is just not the answer to the original problem).
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  8. #8
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    Yeha you're right it wasn't wrong.
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