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I have two questions that pertain to a question that does have quite the complete solution, but I am still having trouble getting over the finish line. The original response is this:

Multivariable limit proof: $\lim\limits_{(x,y)\rightarrow (0,0)}\frac{\left|x\right|^a\left|y\right|^b}{\left|x\right|^c + \left|y\right|^d} = 0$

The question asked to me was to "determine for what positive values $a,b,c,d$ the function is continuous at $0$". The function being:

$$\lim_{(x,y)\rightarrow (0,0)}\frac{\left|x\right|^a\left|y\right|^b}{\left|x\right|^c + \left|y\right|^d} = 0$$

1) My first question revolves around establishing some "possible" conditions for my $a,b,c,d$. To do this I have to try and get the continuity to work. We can see that the function is always positive so it is a candidate to use the squeeze theorem on. There are two cases

i) if $|x|^{c} \leq |y|^{d}$

ii) if $|y|^{d} \leq |x|^{c}$

Doing the first case (as the second will be a mirror of this)

I arrive at

$$ \frac{\left|x\right|^a\left|y\right|^b}{\left|x\right|^c + \left|y\right|^d} \leq \frac{|x|^{a}|y|^{b}}{|x|^{c}} \\ \Rightarrow\ |x|^{a-c}|y|^{b}$$

So if $a \geq c$ then takng the limit of this will go to $0$ and I can squeeze my original function. Same idea would occur for the other case. So I would be left with a scenario where if $a \geq c$ and $b \geq d$ then $f(x,y)$ is continuous at $0$.

I don't feel that this is enough and I'm missing more to the conclusion, but I'm at a loss of how to proceed.

2) My version of the question does not have the added assumption of $\frac{a}{c} + \frac{b}{d} > 1$. To which I believe this is what I have to arrive at as my final conclusion. At this point in the textbook due to the location of the question I don't think I'm not allowed to use any differential calculus, even though in Brian M. Scott's solution he did use single variable calculus which is probably allowed here as well. But from what I know from my studies to this point I don't see how I could arrive at a conclusion of $\frac{a}{c} + \frac{b}{d} > 1$ if that is the end game.

Could I get some help on how to proceed/approach this?

D.C. the III
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1 Answers1

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You wrote that: if $\ |y|^d \leqslant |x|^c \ $ , then: $\ \dfrac{|x|^a|y|^b}{|x|^c + |y|^d} \leqslant |x|^{a-c}|y|^b$

We can go one step further, because $|y|\leqslant |x|^{\frac{c}{d}}$ . So:

$\ \dfrac{|x|^a|y|^b}{|x|^c + |y|^d} \leqslant |x|^{a-c}|y|^b \leqslant |x|^{a-c} |x|^{\frac{bc}{d}} = |x|^{c\left(\frac{a}{c}+\frac{b}{d}-1\right)}$

Same idea when $|x|^c \leqslant |y|^d$.

We can now conclude that, if $\ \dfrac{a}{c}+\dfrac{b}{d}-1 >0\ $ , then $\displaystyle \lim_{(x,y)\rightarrow (0,0)} \dfrac{|x|^a|y|^b}{|x|^c+|y|^d}=0$

If $\ \dfrac{a}{c}+\dfrac{b}{d}-1 \leqslant 0 \ $ , then, if $\ |x|^c=|y|^d \ $, then:

$\ \dfrac{|x|^a|y|^b}{|x|^c + |y|^d} =\dfrac{1}{2} |x|^{a-c}|y|^b =\dfrac{1}{2} |x|^{a-c} |x|^{\frac{bc}{d}} = \dfrac{1}{2}|x|^{c\left(\frac{a}{c}+\frac{b}{d}-1\right)}$

And it's not difficult to conclude.

perroquet
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  • Actually I used the first case, which means I used $|x|^{c} \leq |y|^{d}$ as my assumption and doing the math I'm not getting it to work out how it should. Using the first assumption I would have $\frac{\left|x\right|^a\left|y\right|^b}{\left|x\right|^c + \left|y\right|^d} \leq \frac{|x|^{a}|y|^{b}}{|y|^{d}} \leq \frac{|x|^{a}|y|^{b}}{|x|^{c}} = |x|^{a-c}|y|^{b}$. But then manipulating that last expression using my assumption means $|x|^{\frac{c}{d}} \leq |y|$ and that causes problems to bound the whole expression – D.C. the III May 24 '21 at 22:29
  • @dc3rd The inequality $\dfrac{|x|^a|y|^b}{|x|^c+|y|^d} \leqslant \dfrac{|x|^a|y|^b}{|x|^c}$ is always true when $x\neq 0$. It's a good idea to use it when $|x|^c>|y|^d$. But you won't be able to conclude if you want to use it when $|x|^c<|y|^d$. The inequality $\dfrac{|x|^a|y|^b}{|x|^c+|y|^d} \leqslant \dfrac{|x|^a|y|^b}{|y|^d}$ is always true when $y\neq 0$. It's a good idea to use it when $|x|^c<|y|^d$. But you won't be able to conclude if you want to use it when $|x|^c>|y|^d$. – perroquet May 25 '21 at 21:06
  • I had a feeling that's what you were getting at. Thanks for the help – D.C. the III May 25 '21 at 23:29