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Multivariable Limit, is my reasoning correct?

  1. Sep 26, 2007 #1
    1. The problem statement, all variables and given/known data

    [tex]\lim_{(x,y) \rightarrow (0,0)} \frac{x^2+sin^2 y}{2x^2+y^2}[/tex]


    2. Relevant equations


    3. The attempt at a solution

    Since I can't evaluate the limit directly and I can't see a way to get a 0 on the top in order to get 2 different limits I split up the limit into these 2:
    [tex]\lim_{(x,y) \rightarrow (0,0)} \{\frac{x^2}{x^2+y^2}+\frac{sin^2}{x^2+y^2}\}=\lim_{(x,y) \rightarrow (0,0)} \frac{x^2}{2x^2+y^2}+\lim_{(x,y) \rightarrow (0,0)} \frac{sin^2 y}{2x^2+y^2}[/tex]

    and I work on the one on the left.

    [tex]\lim_{\substack{x=0\\y\rightarrow 0}} \frac{x^2}{2x^2+y^2}=\lim_{y \rightarrow 0} \frac{0}{y^2}=\lim_{y \rightarrow 0} 0 = 0[/tex]

    [tex]\lim_{\substack{y=x\\x\rightarrow 0}} \frac{x^2}{2x^2+y^2}=\lim_{x \rightarrow 0} \frac{x^2}{3x^2}=\lim_{x \rightarrow 0} \frac{1}{3}=\frac{1}{3}\neq0[/tex]

    and since the left limit doesn't exist it doesn't matter if the right limit exists since the sum of a nonexisting limit and something else won't exist either, right?
     
  2. jcsd
  3. Sep 26, 2007 #2

    Avodyne

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    Science Advisor

    Not necessarily; the right limit might not exist, but the nonexistence might cancel between the two terms. For example, if the problem was (x^2+y^2)/(x^2+y^2), the limit is obviously 1, but the limits of x^2/(x^2+y^2) and y^2/(x^2+y^2) do not individually exist.

    But in your case, when y is small, what happens to sin(y)? Just do your two examples with the full numerator.
     
    Last edited: Sep 26, 2007
  4. Sep 26, 2007 #3
    oo because values near 0, sin(y)=y so it would be about the same method of evaluating!

    thanks alot!
     
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