Limits and Continuity

[KBS4000]

Hello, I have a problem with limits and continuity while some being simple for me, others not so much.

The question is What conditions must the non-negative integers m,n, and p satisfy to guarantee that the limit exists prove your answer

limit (x,y) ->(0,0)

(x^m*yⁿ)/ (x² + y²)^p

I got a couple of hints like changing x² + y2 to r² and my teacher wrote this down in red ink, but still not sure what he means

x^m+n/ (2x²)^p

 

[KBS4000]

Sorry about the spamming. I understand you don't give answers, but rather help so I would just like another opinion on the matter because no matter what I do I always come to the conclusion that the limit does not exists since x and y both equal 0

 

[HallsofIvy]

In order that a limit in two dimensions exist, we must get the same thing as we approach along any path. Your teacher is suggesting that you try approaching (0, 0) along the line y= x. On that line, \(\displaystyle x^ny^m= x^nx^m= x^{m+n}\) and \(\displaystyle x^2+ y^2= x^2+ x^2= 2x^2\). So you have \(\displaystyle \frac{x^{m+n}}{(2x^2)^p}= \frac{x^{m+n}}{2^px^{2p}}= \frac{1}{2^p}x^{m+ n- 2p}\).

Now consider the cases m+ n- 2p> 0, m+ n- 2p= 0, m+ n- 2p< 0.

(Remember I said "we must get the same thing as we approach along any path". So you will also want to consider what happens as you approach (0, 0) along the line y= 0 and along the line x= 0.)

 

[KBS4000]

well I may be completely wrong but with

2pxm+n−2p ​

.
if x is 0 wouldn't it be 0/2P

I looked at the three cases you presented and I'm not seeing it. I thought maybe that if equal to 0 m+n= 2p or m+n<2p