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?
Check if below limits exist $\lim_{(x,y,z) \to (0,0,0)} \left( {\frac{{2{x^2}y - x{z^2}}}{{y^2 - xz}}} \right)$?
1
$\begingroup$
calculus
real-analysis
limits
-
2This 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? – 2012-07-24
-
0If 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. – 2012-07-24
-
0@DonAntonio. Do you want to delete my post? – 2012-07-24
-
0Not 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... – 2012-07-24
-
0*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. – 2012-07-24
-
0@Teddy. OK, I'll take that into account, thanks. – 2012-07-24
1 Answers
2
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.
-
0Many thanks! is good! – 2012-07-24
-
0@Teddy is this unique way to solve such problem? – 2012-07-24
-
0@Teddy. What criteria do you use to calculate the sequence? – 2012-07-24
-
0@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. – 2012-07-24
-
0@Teddy. Very ingenious, good point. – 2012-07-24
-
0wonderful solution. +1 – 2012-07-24