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Evaluating undefined limits

  1. Sep 1, 2012 #1
    1. Evaluate



    2. lim χ→0 [((χ+1)^1/3) -1] / χ



    3. The hint is [a^2 + b^2 = (a+b)(a^2 + ab + b^2)].
    But i don't get it how can that help me solve this because it is a cube root and not to the power of 3.


    Without thinking about the hint, I have attempted it and i think it is undefined for the function.
    Say f(x) is the function, so f(0) is undefined making the limit as x approaches 0 undefined as well.
    is it correct??
     
    Last edited: Sep 1, 2012
  2. jcsd
  3. Sep 1, 2012 #2

    Ray Vickson

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    The hint is incorrect: a^2 + b^2 has 'a' and 'b' to the second power, while the right-hand-side has them to the third power. A correct hint would be a^3 - b^3 = (a-b)(a^2 + ab + b^2).

    Think of setting y = (x+1)^(1/3), so the numerator of your expression is y-1. Can you see how to relate y-1 to y^3 - 1? Can you see how to write the denominator (x) in terms of y? Can you see what happens to y when x --> 0?

    RGV
     
  4. Sep 1, 2012 #3

    micromass

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    That's one possibility and it works fine.
    But I prefer to multiply numerator and denominator with [itex]a^2+ab+b^2[/itex] for suitable a and b. What do you get if you use the (correct version of the) hint?
     
  5. Sep 10, 2012 #4
    sorry I was wrong, the hint is: a^3 - b^3 = (a-b)(a^2 + ab + b^2)

    Ok, I did L'hopitals rule to derive the formula, until I am able to get a value from substituting zero in place of x. My answer came to 1/3. But still the hint had nothing to do with it.

    Are you saying for me to do the opposite operation??
     
  6. Sep 10, 2012 #5

    SammyS

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    Here is how the hint might be used:

    The numerator [itex]\sqrt[3]{x+1}-1[/itex]. If you could get an equivalent expression with a numerator of [itex]\displaystyle \left(\sqrt[3]{x+1}\right)^3-1^3\,,[/itex] that would allow major simplification including cancelling a factor that goes to zero in the numerator & denominator.

    Use the hint to find out what you need to multiply [itex]\sqrt[3]{x+1}-1[/itex] by in order to get [itex]\displaystyle \left(\sqrt[3]{x+1}\right)^3-1^3\ .[/itex]

    By the way, welcome to PF!
     
  7. Sep 13, 2012 #6
    so then I would have to multiply it with ([itex]\sqrt[3]{x+1}[/itex])^2 + [itex]\sqrt[3]{x+1}[/itex] +1
    and then do the same with the numerator??
    is it??

    and afterwards derive it???
     
    Last edited: Sep 13, 2012
  8. Sep 13, 2012 #7

    SammyS

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    No, you should not need to use L'Hôpital's rule.

    Just multiply the numerator & denominator by [itex]\displaystyle \left(\sqrt[3]{x+1\ }\right)^2+\sqrt[3]{x+1\ }+1\ .[/itex] Then simplify.
     
  9. Sep 13, 2012 #8
    oh YES

    solved!

    other than simplifying we can just substitute 0, right?
     
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