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Thread: Limit with fractions

  1. #1
    Jul 2012
    United States

    Limit with fractions

    I've been stuck on this problem for a few days and was hoping somebody could point me in the right direction.

    $\displaystyle lim{yto4}\frac{\frac{1}{y}}{y-4}-\frac{\frac{1}{4}}{y-4}$

    I try to flip the denominator (y-4) and multiply across $\displaystyle \frac{1}{y}*{\frac{1}{y-4}-\frac{1}{4}*{\frac{1}{y-4}$ which yields $\displaystyle \frac{1}{y^2-4y}-\frac{1}{4y-16}$

    And then I multiply one side by 4/4 and the other by y/y which gives $\displaystyle \frac{4}{4y^2-16y}-\frac{y}{4y^2-16y}$ ... When I go to evaluate the limit here I get 0/0 which I already know is incorrect. Even if I simplify the 4's further I get a wrong answer, and I guess at this point I'm just not sure if I'm going in the right direction.

    Sorry about the LaTeX being a little sloppy, I kind of typed this in a hurry before class. The top should be "lim as y approaches 4"
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  2. #2
    Member anonimnystefy's Avatar
    Jul 2011

    Re: Limit with fractions


    $\displaystyle \lim_{y\to 4}\frac{\frac{1}{y}}{y-4}-\frac{\frac{1}{4}}{y-4}=$

    $\displaystyle \lim_{y\to 4}\frac{\frac{1}{y}-\frac{1}{4}}{y-4}=$

    $\displaystyle \lim_{y\to 4}\frac{\frac{4-y}{4y}}{y-4}=$

    $\displaystyle \lim_{y\to 4}\frac{4-y}{4y(y-4)}=$

    $\displaystyle \lim_{y\to 4}-\frac{1}{4y}=$

    $\displaystyle -\frac{1}{4\cdot 4}=-\frac{1}{16}$

    Hope this helps. If you have any more questions, feel free to ask.
    Thanks from zikcau25
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  3. #3
    Senior Member MaxJasper's Avatar
    Aug 2012

    Question Re: Limit with fractions

    Last edited by MaxJasper; Sep 4th 2012 at 10:43 AM.
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  4. #4
    Junior Member
    Apr 2012
    Rose Belle, Mauritius

    Re: Limit with fractions

    That elaborate formula isnít in its simplest form. It includes an impurity that you need to pinpoint and eliminate. It should have been taught for people to know about it or maybe it demands only a little attention to discover them.

    In my first post post1 on Trigonometric identities problem, for the first time I manage to identify an impurity and coincidentally I just find the same one in your problem.

    Well, now the matter seems a bit too serious to be ignored. The algebraic impurity that I called is in fact an algebraic identity as well as a general fact:

    ďAny sum of terms, monomials to polynomials, divided by its negative sum always equal to -1.Ē

    for example : (x+1)/(-x-1), (4-y)/(y+4), (x-1+2-3+y+z)/(1-2+3-y-z-x)etc... all equals -1.
    (Numerically same,numerator and denominator having same magnitude, but differs in sign - Hence ratio of similar numbers make 1 and (-) is conserved)

    Therefore, dealing from this aspect:

    Limit with fractions-solving.jpg
    From this step on, you can proceed on with the problem normally.

    Hope this approach is worthy of some importance.
    Last edited by zikcau25; Sep 4th 2012 at 01:44 PM.
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