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Does limit exist? (i have test tomorrow morning!)

  1. Oct 22, 2013 #1
    1. The problem statement, all variables and given/known data
    Does the limit of (2y*x^2)/(4x^2-3y) exist as (x,y) -> (1,1)? If so what is the limit?


    3. The attempt at a solution
    So at first I thought I could just plug (1,1) into the equation and get two. My study buddy said that you have to make sure it approaches the same thing from every direction else the limit DNE. If it approaches the limit from x=0 then it equals zero, and same with y. but if you take the limit with y=x then it comes out to two.

    Are we overthinking this? Can't you just plug it in and ignore the "approaching from x=0" since x can't approach from zero because it goes to 1?
     
  2. jcsd
  3. Oct 22, 2013 #2

    jbunniii

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    Hang on, I had typos in my previous post. Will resubmit with corrections.
     
  4. Oct 22, 2013 #3

    jbunniii

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    Your buddy is right that if you find different limits when approaching via different paths, then the limit does not exist.

    If you put ##y = x^2## (I assume you meant this, not ##y=x##), then you get ##2x^4/x^2 = 2x^2##, which indeed approaches ##2## as ##x \rightarrow 1##.

    However, "approaching via ##x=0##" makes no sense in this problem because then you won't approach the point ##(1,1)##.
     
  5. Oct 22, 2013 #4

    jbunniii

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    Here's another approach to the problem. Note that there is no problem with the denominator as you approach ##(1,1)##. So the quotient is well defined in the vicinity of that point. The numerator and denominator are continuous functions of ##x## and ##y## (proof?), so the quotient is also continuous in a neighborhood of ##(1,1)##. What can you conclude?
     
  6. Oct 23, 2013 #5

    HallsofIvy

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    Tell your study budy he is "overthinking" this. The only time there is a "problem" with a limit so that you need to check "along different paths" is if taking numerator and denominator separately go to 0 so that you get the "indeterminate form" 0/0. Here, that does not happen. In consequence, "the quotient is continuous in a neighorhood of (1, 1)" as jbunniii says.
     
  7. Oct 23, 2013 #6

    Borek

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    Somehow I miss mentioning of one-side limits. I mean - you can have a one side limit(s) in the point, but no limit. Am I right?
     
  8. Oct 23, 2013 #7

    jbunniii

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    The notion of one-sided limits is only applicable to functions of one variable. Here we have a function of two variables, so instead of only two possible directions to approach a point (from the left and the right) we have infinitely many paths. If you can find even two paths which disagree, then the limit doesn't exist. However, this can never happen if the function is continuous at the point of interest.
     
    Last edited: Oct 23, 2013
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