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I am having trouble deriving the correct answer for this problem. I am trying to find the limit of the following problem:

$$\lim _{(x,y) \rightarrow (0,0)} \frac{-3xy}{3x^2+y^2}$$

Here is my work:

$$\frac {-3xy}{3x^2+y^2}= \frac{(-3x)(mx)}{(3x^2+(mx)^2)} = \frac{-3mx^2}{(x^2(3+m^2))} = \frac{-3m}{(3+m^2)} = \frac{-3}{(3+m)}$$

I keep getting that the limit DNE because it is dependent on m, but that is not the correct answer for some reason? Any help would be great, thanks!

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    Where is your work ? – 2017-02-13
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    Sorry, I was adding it. I just edited it on there – 2017-02-13
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    The last equality is not correct, the correct one should be $$\frac{-3m}{3+m^2}=\frac{-3}{\frac{3}{m}+m}.$$ – 2017-02-13

4 Answers 4

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Using polar substitution your limit becomes $$\lim _{r \to 0} \frac{-3}{2}\frac{r^2\sin(2\theta)}{3r^2\cos^2(\theta)+r^2\sin^2(\theta)}$$ Clearly we can factor out $r^2$ to get $$\lim _{r \to 0} \frac{-3}{2}\frac{\sin(2\theta)}{3\cos^2(\theta)+\sin^2(\theta)}$$
We now see that the limit does not depend on $r$ at all in the sense that it is indeterminent; we can make it anything by changing $\theta$. You are definitely right in that the limit is path dependent.

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    Hm okay then it must be the homework software, because it is saying that the answer is not DNE. – 2017-02-13
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Try the path $x=y$ to the origin. Then try the $x$ axis as a path.

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If we try the x=y path we are asking for $\lim_{x=0}\frac{-3x^2}{4x^2}=-\frac{3}{4}.$

If we try the x-axis path (set y=0) we get 0 obviously.

So you are right that the limit is indeterminate.

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    Hm okay then it must be the homework software, because it is saying that the answer is not DNE. – 2017-02-13
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Tracing along different paths gives difference limits, and thus the limit cannot be determined.