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Math Help - Confusing Limit problem

  1. #1
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    Confusing Limit problem

    Hi, I'm confused on how to figure out the problem \lim_{x \to 27} \frac{x-27}{x^\frac{1}{3}-3}. I have the problem worked out, but I am having difficulty understanding how to get from step one to step 2:
    \lim_{x \to 27} \frac {(x^\frac{1}{3})^3-(3)^3}{x^\frac{1}{3}-3}

    to

    \lim_{x \to 27}\frac {(x^\frac{1}{3}-3)(x^\frac{2}{3}+3x^\frac{1}{3}+9)}{(x^\frac{1}{3}-3)}

    Thanks for the help in advance.
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  2. #2
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    It was done using the difference of 2 cubes.

    a^3- b^3 = (a - b)(a^2 + ab + b^2)
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  3. #3
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    It is just factorisation.

    For example,

    If f(x)=x^{3}-27

    Since f(3)=0, by factor theorem, (x-3) is a factor.

    So f(x)=(x-3)(ax^{2}+bx+c)

    Do either long division or comparison to get the other quadratic factor, you will get f(x)=(x-3)(x^{2}+3x+9).

    In your limit problem, my x is your x^{\frac{1}{3}}
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  4. #4
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    (x^\frac{1}{3})^3-(3)^3

    You can do synthetic division:

    (x^\frac{1}{3})^3+0*x^\frac{2}{3}+0*x^\frac{1}{3}-27

    x^\frac{1}{3}-3=0
    x^\frac{1}{3}=3

    So:

    3 | 1 0 0 -27
    _ |___3 9 27
    ___ 1 3 9 0

    hmm don't know how to do this in latex, nor can i get it to line up . . . sorry.


    So you lower the power thing and:

    1*x^\frac{2}{3}+3*x^\frac{1}{3}+9

    hence:
    (x^\frac{1}{3}-3)(x^\frac{2}{3}+3x^\frac{1}{3}+9)
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  5. #5
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    A slightly different way to do this problem would be to "rationalize" the denominator.

    Since a^3- b^3= (a- b)(a^2+ ab+ b^2), [tex](x^{1/3}- 3)(x^{2/3}+ 3x^{1/3}+ 9)= x- 27. Multiplying both numerator and denominator by x^{2/3}+ 3x^{1/3}+ 9 gives
    \frac{(x- 27)(x^{2/3}+ 3x^{1/3}+ 9)}{x- 27}= x^{2/3}+ 3x^{1/3}+ 9. Now you can let x= 27 in that.
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  6. #6
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    Quote Originally Posted by cynlix View Post

    Hi, I'm confused on how to figure out the problem \lim_{x \to 27} \frac{x-27}{x^\frac{1}{3}-3}.
    put t=x^{\frac13} and the limit becomes \underset{t\to 3}{\mathop{\lim }}\,\frac{t^{3}-27}{t-3}, which you know how to solve.
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