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Homework Help: Limits with two variables, explain why limit DNE

  1. Mar 12, 2012 #1
    1. The problem statement, all variables and given/known data
    (Sorry, having problems with math symbols)

    lim f(x,y) [(x^4)y]/(x^8+y^4)

    2. Relevant equations

    Compare limits when

    a.) y=x
    b.) y=x^4

    3. The attempt at a solution

    The solutions are

    a.) limit approaches 0
    b.) limit approaches 1

    I think I understand in order for a limit to exist, it must approach the same value regardless of which direction we approach it from. So I think that the limit DNE here because there are two different values approached. However, not really sure how to substitute values into the equation without getting zero for both a and b.
  2. jcsd
  3. Mar 12, 2012 #2
    You are correct about the limit not existing.

    To evaluate the limits, you are given formulas for what y is in terms of x. You can simply plug these equations in and eliminate a variable and solve the limit. Since Both equations approach the same point, the variable that is still left approaches 0. You are effectively making a 2-d problem into a 1-d problem.

    So, if [itex]y=x[/itex], [itex]\displaystyle\lim_{(x,y) \rightarrow (0,0)} \displaystyle\frac{x^{4}y}{x^{8}+y^{4}} = \displaystyle\lim_{x \rightarrow 0} \displaystyle\frac{x^{5}}{x^8+x^4}[/itex]. What do you get when you solve this limit?

    Similarly, for [itex]y=x^{4}[/itex], [itex]\displaystyle\lim_{(x,y) \rightarrow (0,0)} \displaystyle\frac{x^{4}y}{x^{8}+y^{4}} = \displaystyle\lim_{x \rightarrow 0} \displaystyle\frac{x^{8}}{x^{8}+(x^{4})^{4}}[/itex]. What is the limit of this?
  4. Mar 12, 2012 #3
    (I am assuming we are evaluating the function at the point (0,0), if this is correct terminology?)

    When y=x, the x^8 is evaluated at 0, but what happens to the remaining fraction x^5/x^4. Is that simply "x", which is then evaluated again at 0?

    When y=x^4, the x^8 in the denominator again becomes 0, and a fraction remains that is x^8/x^8 which is 1? Thanks for your reply scurty

    But perhaps this is not simply evaluating a function at point (a)...
  5. Mar 12, 2012 #4
    Yes, that's exactly what you are doing. However, you are approaching the point (0,0) from two different directions in parts a) and b). Because the limit values are not the same when approached from ALL directions, the limit does not exist.

    Think of this problem when only considering one variable (i.e. the real plane). Say you want to evaluate [itex]\displaystyle\lim_{x \rightarrow 0} \frac{1}{x}[/itex]. To solve this you need to make sure [itex]\displaystyle\lim_{x \rightarrow 0^{-}} \frac{1}{x} = \displaystyle\lim_{x \rightarrow 0^{+}} \frac{1}{x}[/itex]. Only then does the limit value exist! (in this case the limit does not exist)

    This is no different except we are considering two variables. To make the calculation easier we substitute for variables to reduce it to one.

    Not quite. Try factoring out the greatest common factor out of the denominator and see what you are left with!

    Try the same method I suggested above. We are taking the limit as x approaches 0, not as [itex]x^{4}[/itex] approaches 0.

    Does that make more sense hopefully?
  6. Mar 12, 2012 #5
    Thanks for mentioning we are reducing the problem from two dimensions to one dimension.. and for providing that example. Jarring my memory of calc 1 and now I see "of course that's what we were doing back then!"

    Let's see how I did...


    (x^4/x^4)[x/(x^4+1)]→(x=0)→(0/0+1) = 0


    (x^8/x^8)[1/(1+x^8)] →(x=0)→(1/1+0) = 1

    limit DNE
  7. Mar 12, 2012 #6
    Exactly! Nice work! Sometimes the tricky problems are coming up with two different equations to approach the point at to show the limit doesn't exist by yourself. In this problem they gave you the two so it wasn't so bad.
  8. Mar 12, 2012 #7
    Thanks Scurty! This forum is awesome!
  9. Mar 12, 2012 #8


    Staff: Mentor

    I'm assuming that what you mean by "the function," you mean ## \frac{x^4y}{x^8 + y^4}##.

    If so, you cannot evaluate this function at the point (0, 0), because this function is undefined there. That's the reason that you are asked to evaluate the limit as (x, y) → (0, 0).

    Along each the specified paths, the function simplifies to a different function that involves only x, and you can take the limit as x → 0.

    Note that for both limits, you cannot simply evaluate the limit expression at x = 0, as both limit expressions are undefined at x = 0. Using the properties of limits, however, you can evaluate both limits.

    As you have found, on each of these paths a limit exists, but it is not the same limit for both paths.

    No it isn't. The function in the original limit is undefined at the point (0, 0). In addition, along each of the two paths, the simplfied limit expression was undefined at x = 0.
  10. Mar 12, 2012 #9
    Sorry, I misread his text. I read it as evaluating the limit of the function at the point (0,0), he left out the limit part though.
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