Evaluate limit of $|x|^y$ as $(x,y ) \to (0,0)$

Question

I have no idea where to start, I tried transforming it into $e^{yln|x|}$ but have I dont know what should I do next.

this limit does not exist. try finding two sequences giving different limits — mm-aops, May 27 '14 at 12:45
Recently, in the Mathematics Educators stackexchange question Multivariable limits, I discussed the fact that $x^y$ has no limit as $(x,y) \rightarrow (0,0)$ in the first quadrant. Note that this is a stronger result than what you're asking. — Dave L. Renfro, May 27 '14 at 15:04

score 3 · Answer 1 · answered May 27 '14 at 12:46

3

Hint: Try showing that the limit does not exist. Aprroach $(0,0)$ along the line $x=0$ and along the line $y=0$, respectively.

answered May 27 '14 at 12:46

5xum

123,496
6
128
204

And how do I find the limit of $0^y$ for y->0 – Lugi May 27 '14 at 12:51
$0^y=0$ for any $y>0.$ So, $\lim_{y\to 0^+} 0^y=0.$ – mfl May 27 '14 at 12:59
@user153653 By first calculating $0^y$ for $y\neq 0$. – 5xum May 27 '14 at 12:59

score 1 · Answer 2 · answered May 27 '14 at 12:46

1

That limit is undefined since it depends on how $x,y$ go to zero.

answered May 27 '14 at 12:46

JPi

4,562

Aryan · Answer 3 · 2014-05-27T13:29:44.523

0

As previously stated the limit does not exist.

This can be proved by contradiction. If the limit did exist then the limit value must be the same as starting with $x^y$ and then taking the $x,y$ limits in any order. First take the limit as $y \rightarrow 0$ and then $x \rightarrow 0$. In this case the limit will be 1.

Then take the limit the other way around. Starting with $x^y$ take the limit as $x \rightarrow 0$ and then $y \rightarrow 0$ (ensuring that y>0), this will give the limit as 0.

Therefore the limit of $x^y$ as $(x,y) \rightarrow (0,0)$ does not exist due to non-uniqueness.

edited May 27 '14 at 13:29

answered May 27 '14 at 13:23

Aryan

41

Not Correct! First suppose that $x=1E-12$, then if we let $y=1E-12$ we have $x^y=1$. I say that limit exist and is equal to $1$. – SKMohammadi May 27 '14 at 13:42
Hi MrMath, please edit your message as your post hasn't rendered properly for me to read. – Aryan May 27 '14 at 14:38
Hi MathMan also if you look at your 1st Mathematica Graph notice on the bottom right $x^y$ follows $0$, thereafter $x^y$ quickly jumps to $1$ at the origin. This is because $0^0$ is defined as $1$ in most Maths software including Julia, and we should appreciate that to find a limit the function doesn't need to be defined at $(0,0)$. On that basis the Mathematica graph actually concurs with the previous posts that the limit takes on different values (which you can actually see in your graphs). The function value approaches $0$ on the bottom right $1$ on the top left (first graph). All the best. – Aryan May 27 '14 at 15:10

score 0 · Answer 4 · answered May 27 '14 at 13:29

0

Plot of this function using Mathematica:

enter image description here

answered May 27 '14 at 13:29

SKMohammadi

1,091
8
21

Evaluate limit of $|x|^y$ as $(x,y ) \to (0,0)$

4 Answers4