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Limit (2 variables)

  1. Mar 8, 2017 #1
    1. The problem statement, all variables and given/known data

    2. Relevant equations
    this is the entire syllabus of the test from which i got this question. Kindly mention if this question is out of syllabus

    3. The attempt at a solution
    I don't have the idea to approach this question. In normal limits, i know that graphically it is the converging point of the plot of the function (which will be an arc) when the function approaches the required x value. But here there are two variables. By which variable should i differentiate when using l'hospitals rule here .No idea how to start :(
  2. jcsd
  3. Mar 8, 2017 #2
    This syllabus is for which test ?

    I don't think the limit exists.
  4. Mar 8, 2017 #3

    Why do you think limit doesn't exist?
  5. Mar 8, 2017 #4

    Stephen Tashi

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  6. Mar 8, 2017 #5


    Staff: Mentor

    Thread moved. Questions about limits of functions of more than one variable are typically found in calculus courses, not in precalc courses.
    This doesn't make much sense. The function doesn't approach "a required x value." And the graph of a function may or may not have what you're calling a converging point. For example, ##\lim_{x \to 0}\frac 1 {x^2}## doesn't have a converging point (this limit is ##\infty##).
    L'Hopital's Rule is for limits of the form ##\frac{f(x)}{g(x)}##; i.e., the quotient of single variable functions. If there's a form of L'Hopital's Rule for functions with two or more variables, I've never seen it.

    With limits of functions of two variables, as you have here, if the limit exists, it must be the same along any path to, in this case, (0, 0). Possible paths include along the x-axis, along the y-axis, along the line y = mx, along various curves. If by approaching along two different paths you get a different limiting value, then the limit does not exist.
  7. Mar 8, 2017 #6


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    This looks like an example getting at cases where Clairaut's theorem (or Schwarz's theorem) cannot be applied, and you'd have a have a non-symmetric Hessian, which is problematic. Put differently, this looks to be illustrating a case where second partials exist but they are not continuous -- at (0,0).

    These rarely come up in practice, but you need to be aware of them. (For a non-horrific function, the tell tale sign is the special handling with the if statement to avoid 0 / 0.)
  8. Mar 9, 2017 #7
  9. Mar 9, 2017 #8
    Sorry I made a typing mistake there. What I meant is as x approaches a required value(say a), (from right or left) if the function approaches a specific value (y) (or ±infinity) independent of direction in which x approaches a, then the limit of the function as x goes to a is y.

    If y is dependent on the direction of x->a , then the limit does not exist.

    Am i right now ?
  10. Mar 9, 2017 #9


    Staff: Mentor

    This isn't too far off if you're talking about the limit of a function of one variable.
  11. Mar 9, 2017 #10
    Like in single variable you approach a limit from two sides left and right, in two variable limit you approach the limit from all the possible directions like from like ##y = x##, ##y= -x## etc.

    Limit from both sides should converge to same point in single variable for the limit to exist, same applies to multivariable in which limit from all the direction should converge at same point.

    We have done only a little bit of 2 variable limit in electrodynamics course. So I don't know much about these but it looks like this limit does not exist.
  12. Mar 9, 2017 #11

    Stephen Tashi

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    In the case of multivariable limits, the same does not apply - in the sense of "if...then...".

    In general, True: 1) if ##lim_{(x,y) \rightarrow (a,b)} f(x,y) ## exists and is equal to ##L## then the limit for any direction exists and is also equal to ##L##

    In general, False: 2) If the limit from any direction exists and is equal to ##L## then ##lim_{(x,y) \rightarrow(a,b)} f(x,y)## exists and is also equal to ##L##.

    Yes, but the limit taken along any straight line ## y = mx## exists and is equal to zero, which illustrates that 2) is false.

    See example 14.2.1 of https://www.whitman.edu/mathematics/calculus_online/section14.02.html
  13. Mar 9, 2017 #12
    Thanks I already told I don't know much about multivariable limits.
    Mod note: deleted some work in this post that the OP needs to do
    Last edited by a moderator: Mar 9, 2017
  14. Mar 12, 2017 #13
    I guess i got it now.
    Suppose i am taking the direction of the curve y = 2x2,
    Then the limit is 2/5

    And if i take y = 3x2,
    the limit is 3/10

    So the limit does not exist.

    Am i right ?
  15. Mar 12, 2017 #14


    Staff: Mentor

    Yes. Since you have demonstrated two paths that give different values, the limit does not exist.
  16. Mar 12, 2017 #15
    Yes, because in any metric space, a limit, if it exists, is uniquely determined. [Doesn't generally hold in topological spaces]
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