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Limits and L'Hospital

  1. Mar 25, 2012 #1
    1. The problem statement, all variables and given/known data
    Determine if limit exists:
    ## \left( !x_{,y}^{im}\right) \rightarrow \left( 1,2\right) =\dfrac {xy-2x-y+2} {x^{2}-2x+y^{2}-4y+5} ##

    Above is just lim (x,y)-->(1,2)
    2. Relevant equations



    3. The attempt at a solution
    ## c_{y}x=a,y\rightarrow b ##
    ## \lim _{\left( 1,y\right) \rightarrow \left( 1,2\right) }\dfrac {y-2-y+2} {1-2+y^{2}-4y+5}=\dfrac {0} {y^{2}-4y+4} ##
    Why wasn't L'Hospital used?
     
  2. jcsd
  3. Mar 25, 2012 #2

    lanedance

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    when would you normally apply L'hopital?

    Is always worth having a think about the behaviour of the numerator and denominator as you approach the limit point. What do they do as you get close to (1,2)?
     
  4. Mar 25, 2012 #3
    I apply L'Hospital, when I have an indeterminate equation. Namely,0/0, (infinity/infinity), (-infinity/ -infinity), and their combinations.

    As I get close to (1,2) the function becomes 0/0.
     
  5. Mar 25, 2012 #4

    HallsofIvy

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    How do you apply L'Hopital's rule to a function of two variables?
     
  6. Mar 25, 2012 #5
    When my function is indeterminate, I find the derivative of the numerator and denominator and if necessary consecutive derivatives of each until I can find a limit that is not indeterminate.
     
  7. Mar 25, 2012 #6

    SammyS

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    (It's the limit of an expression, it's not an equation.)

    As long as y ≠ 2 , the expression [itex]\dfrac {0} {y^{2}-4y+4} [/itex] is zero .

    To find the limit that's asked for, [itex]\displaystyle \lim_{(x,y)\to(1,2)}\ \left(\dfrac {xy-2x-y+2} {x^{2}-2x+y^{2}-4y+5}\right)[/itex], do one of the following:
    If the limit doesn't exist, show that the limit depends upon the path taken as (x, y) approaches (1, 2). For this problem it may be enough let (x, y) approach (1, 2) along a line of arbitrary slope (although that is often not sufficient).
    y = m(x-1) + 2 .​
    If the limit does exist, polar coordinates are often helpful. In this case it would be advisable do a coordinate translation first: Letting u = x-1 and v = y-2 .


     
  8. Mar 25, 2012 #7
    I have learned to check for limits through traces

    Find limits while setting some variables constant and the other approaching the given (a,b) a or b.

    Do I need to use L'Hospital in this function?
     
    Last edited: Mar 26, 2012
  9. Mar 26, 2012 #8
    What is "m"?
     
  10. Mar 26, 2012 #9

    SammyS

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    m is the slope of the line.
     
  11. Mar 26, 2012 #10

    HallsofIvy

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    m is the slope of the straight line. By taking different values for m you are approaching along different lines. If the limit depends upon m, you know that different lines will give different results and so the limit, as a function of two variables, does not exist.
     
  12. Mar 26, 2012 #11
    Your approach is confusing to me.

    I found that the limit doesn't exist by finding traces:
    Say,
    Cy{ x=a, y-->b}
    Cx {y=b, x-->a }
    C1{x-->a, y=(b/a)} x (Which is the one I think you are mentioning)

    I found that the limit set by C1 is different than Cx; therefore, the limit does not exist.
    C1 would be:
    ## \lim _{\left( x,2x\right) \rightarrow \left( 1,2\right) }\dfrac {xy-2x-y+2} {X^{2}-2x+y^{2}-4y+5} ## (The capital X above is just x )

    ## \lim _{x\rightarrow 1}\dfrac {2\left( x^{2}-2x+1\right) } {5\left( x^{2}-2x+1\right) }=\dfrac {2} {5}\neq c_{y},Cx ##
     
    Last edited: Mar 26, 2012
  13. Mar 26, 2012 #12

    lanedance

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    Couldn't quite folllow you last piece of working, however if you have shown the limit along 2 distinct linear approaches is different, then you have shown teh limit does not exist

    note the both the lines below pass through (1,2)
    y = 2x
    y = x+1

    The general equation of a line is
    y = mx+c

    you can solve for all lines that pass through (1,2), except for the vertcial one, be noting the above a equation must satisfy
    2 = m1+c
    c = 2-m

    Which gives
    y = mx +2 - m
    y = m(x-1)+2

    If you subsitute this in to the limit, you are effectively testing all linear approaches, excpet for the vertical

    2 of the cases you gave are shown below

    m=2
    y = 2x

    m= 0
    y = 2
     
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