# Check if below limits exist $\lim_{(x,y,z) \to (0,0,0)} \left( {\frac{{2{x^2}y - x{z^2}}}{{y^2 - xz}}} \right)$?

Check if below limits exist $$\lim_{(x,y,z) \to (0,0,0)} \left( {\frac{{2{x^2}y - x{z^2}}}{{y^2 - xz}}} \right).$$ Is there any succession to prove that this limit is not zero?

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This is your third question on the very same theme. Did you learn anything from the answers to the two previous questions? What did you try to attack this one? – Did Jul 24 '12 at 11:23
If after several questions on the same verge you can't show some own effort to attack questions very closely related you're going to get downvoted. – DonAntonio Jul 24 '12 at 11:26
@DonAntonio. Do you want to delete my post? – mathsalomon Jul 24 '12 at 11:30
Not at all, @mathsalomon. What I'd like to see is some ideas from you how to attack this problem. It's understandable you're having a tough time tackling this stuff, but imo it isn't that after several questions on the same subject you don't show some ideas to solve your problem that you may have received from the answers you got. It's just like you're expecting your homework solved for you by others... – DonAntonio Jul 24 '12 at 11:35
Et voilà ! Everything is in place for a fourth question, completey similar to the previous ones, with no hint whatsoever of what the OP tried. – Did Jul 24 '12 at 11:37

## 1 Answer

Taking the sequence $x_n = z_n = \frac{1}{n}$, and $y_n = \frac{1}{n} + \frac{1}{n^2}$, you get $$\lim_{n\to \infty} \frac{\frac{1}{n^3}+\frac{2}{n^4}}{\frac{2}{n^3}+\frac{1}{n^4}} = \frac{1}{2}.$$ You can easily find a sequence for which the limit is $0$, and hence the limit does not exist.

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Many thanks! is good! – mathsalomon Jul 24 '12 at 11:34
@Teddy is this unique way to solve such problem? – dato datuashvili Jul 24 '12 at 11:35
@Teddy. What criteria do you use to calculate the sequence? – mathsalomon Jul 24 '12 at 11:39
@dato The function is not defined along the line $x=y=z$. I found that many times it pays to check a sequence which approaches the limit point, along a curve that is "close" to the curve where the function is undefined. – Teddy Jul 24 '12 at 11:44
@Teddy. Very ingenious, good point. – mathsalomon Jul 24 '12 at 11:46