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Math Help - limits with radicals and with absolute values

  1. #1
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    limits with radicals and with absolute values

    I have two limit questions:

    a) lim x goes to 1 of ((square root (6-x) -2)/ (square root (10-x) -3))
    b) lim x goes to 0 of (2x / (abs(3x-2) - abs(3x+2))

    The first one I tried to multiply by both conjugates but it doesn't result in things canceling like it should. The second one I can't work out the intervals. I graphed it and see the limit is a negative third but I can't see that algebraically.
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  2. #2
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    Re: limits with radicals and with absolute values

    "Things don't cancel like they should" because the numerator is NOT 0 when x= 1. What does that tell you about the limit?

    As for the second, |x| "changes formula" at x= 0. |3x- 2| changes formula when 3x- 2= 0 and |3x+2| changes formula when 3x+ 2= 0. What are the solutions to that? Those values of x divide the real number line into three intervals.
    Last edited by HallsofIvy; October 7th 2012 at 03:05 PM.
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  3. #3
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    Re: limits with radicals and with absolute values

    But the denominator is zero and 1/0 is undefined. So is do I just say the limit is undefined? That doesn't seem like much of a question then.

    I tried setting it up with intervals of less than -2/3, -2/3 < x < 0, 0<x<2/3 and greater than 2/3 but then for the limit to exist all of the intervals should give me the same value. But when I worked it out for x < -2/3 both abs give negative results and the denominator become 4 rather than the -6x I want to get a value of -1/3.
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