Rational Limit Theorem.
For $f(x, y) =\frac{|x|^a|y|^b}{|x|^c+|y|^d}$, with $a, b, c, d$ positive,
$$\lim_{(x,y)→(0,0)}f(x, y) \text{ exists and equals zero}\Leftrightarrow \frac{a}{c}+\frac{b}{d}>1.$$
We will break this down into three parts: first proving one direction by our techniques to show limits don’t exist, then using a famous inequality to help prove the other direction. As a guideline, each proof can be written in two or three lines.
$1.$ Show that if $\frac{a}{c}+\frac{b}{d}≤1$, then the limit does not exist.
For the next two problems, you are free to use the following inequality:
Young’s Theorem. For positive real numbers $w$, $z$ and any $0≤t≤1$,
$$w^tz^{1−t}≤tw+ (1−t)z$$
$2.$ Show that if $\frac{a}{c}+\frac{b}{d}= 1$, where $a, b, c, d$ are all positive, then $f(x, y)≤1$ for all $(x, y)∈\mathbb{R^2}\backslash\{(0,0)\}$.
$3.$ Show that if $\frac{a}{c}+\frac{b}{d}>1$, then $$\lim_{(x,y)→(0,0)}f(x, y) = 0.$$
Before I start to prove this, i'm thinking why $1-3$ together implies
$$\forall a,b,c,d>0,\lim_{(x,y)→(0,0)}\frac{|x|^a|y|^b}{|x|^c+|y|^d} \text{ exists and equals zero}\Leftrightarrow \frac{a}{c}+\frac{b}{d}>1$$
"$1.$" looks like the contrapositive of direction "$\Rightarrow$", actually, "$1.$" implies "$\Rightarrow$", but "$\Rightarrow$" doesn't implies "$1.$", this makes the statement stronger, which is good.
To show "$1.$" implies "$\Rightarrow$"
Let $f(x,y)=\frac{|x|^a|y|^b}{|x|^c+|y|^d}\text{ and },a,b,c,d > 0$, assume "$1.$" we have
$$\frac{a}{c}+\frac{b}{d}\le1\rightarrow \lim_{(x,y)→(0,0)} f(x,y) \text{ not exists}$$
$$\Rightarrow(\frac{a}{c}+\frac{b}{d}\le1 \wedge \lim_{(x,y)→(0,0)} f(x,y)=0)\rightarrow \lim_{(x,y)→(0,0)} f(x,y) \text{ not exists}$$
Since $((a \wedge b)\rightarrow c)\Leftrightarrow(a\rightarrow(\neg b \vee c))$ we have:
$$\Leftrightarrow\frac{a}{c}+\frac{b}{d}\le1\rightarrow (\lim_{(x,y)→(0,0)} f(x,y) \text{ not exists} \vee \lim_{(x,y)→(0,0)} f(x,y)\neq0)$$
Which is the contrapositive of "$\Rightarrow$", so this make sense $\dots$ logically. $\tag*{$\square$}$
Then I suppose "$2.$" and "$3.$" together should implies direction "$\Leftarrow$".
"$2.$" states the following:
$$\forall a,b,c,d>0, \frac{a}{c}+\frac{b}{d}=1\rightarrow \forall (x,y)\in\mathbb{R^2}\backslash\{(0,0)\},f(x,y)\le 1$$
"$3.$" says that:
$$\forall a,b,c,d>0,\frac{a}{c}+\frac{b}{d}>1\rightarrow\lim_{(x,y)→(0,0)}f(x, y) = 0.$$
And together should implies "$\Rightarrow$":
$$\forall a,b,c,d>0,\frac{a}{c}+\frac{b}{d}>1\rightarrow\lim_{(x,y)→(0,0)} f(x,y) \text{ exists} \wedge \lim_{(x,y)→(0,0)} f(x,y)=0$$
Proof.
Assume "$3.$" have:
$$\forall a,b,c,d>0,\frac{a}{c}+\frac{b}{d}>1\rightarrow\lim_{(x,y)→(0,0)}f(x, y) = 0.$$
Since $$\lim_{(x,y)→(0,0)} f(x,y)=0\rightarrow \lim_{(x,y)→(0,0)} f(x,y) \text{ exists}$$
Directly implies "$\Rightarrow$"
$$\forall a,b,c,d>0,\frac{a}{c}+\frac{b}{d}>1\rightarrow\lim_{(x,y)→(0,0)} f(x,y) \text{ exists} \wedge \lim_{(x,y)→(0,0)} f(x,y)=0\tag*{$\square$}$$
$1.$
$$\text{WTS }\forall a,b,c,d>0,\frac{a}{c}+\frac{b}{d}\le1\rightarrow \lim_{(x,y)→(0,0)} f(x,y) \text{ not exists}$$
Proof.
Let $a,b,c,d\in\mathbb(0,\infty)\cap{\mathbb{R}}, S=\mathbb{R^2}\backslash\{(0,0)\}$
Assume
$$\frac{a}{c}+\frac{b}{d}\le1$$
Show the limit does not exist
Let $x=t^\frac{1}{c}, y=mt^\frac{1}{d}$ where $m\ge0$, we have the following
$$\lim_{(x,y)→(0,0)}f(x,y)=\lim_{t→0}\frac{|t^\frac{1}{c}|^a|mt^\frac{1}{d}|^b}{|t^\frac{1}{c}|^c+|mt^\frac{1}{d}|^d}$$
Try approch $t$ from right side, so everything is positive, then we have:
$$\lim_{t→0^+}\frac{mt^{\frac{a}{c}+\frac{b}{d}}}{(m+1)t}
=\lim_{t→0^+}\frac{1}{m+1}t^{\frac{a}{c}+\frac{b}{d}-1}$$
Consider two cases:
Case 1:$\frac{a}{c}+\frac{b}{d}=1$
Have $$\lim_{t→0^+}\frac{1}{m+1}t^{0}=\frac{1}{m+1}$$
The limit depend on the value of $m$, that implies limit d.n.e
Case 2:$\frac{a}{c}+\frac{b}{d}<1$
Have $\frac{a}{c}+\frac{b}{d}-1$ is negative, implies limit diviges, that
$$\lim_{t→0^+}\frac{1}{m+1}t^{\frac{a}{c}+\frac{b}{d}-1}=\infty$$
Therefore in both cases limit d.n.e$\tag*{$\square$}$
$2.$
$$\text{WTS }\forall a,b,c,d>0, \frac{a}{c}+\frac{b}{d}=1\rightarrow \forall (x,y)\in\mathbb{R^2}\backslash\{(0,0)\},f(x,y)=\frac{|x|^a|y|^b}{|x|^c+|y|^d}\le 1$$
Maybe I can use Young's Theorem here
$$\forall w,z\in\mathbb{R},t\in[0,1]\cap\mathbb{R}, w^tz^{1−t}≤tw+ (1−t)z$$
Let $p=\frac{a}{c}>0,q=\frac{b}{d}>0,r=|x|^c,s=|y|^d$, have
$$\frac{|x|^a|y|^b}{|x|^c+|y|^d}=\frac{(|x|^c)^{\frac{a}{c}}(|y|^d)^{\frac{b}{d}}}{|x|^c+|y|^d}=\frac{r^ps^q}{r+s}=\frac{r^ps^{1-p}}{r+s}s^{p+q-1}=\frac{r^ps^{1-p}}{r+s}$$
$$\le\frac{pr+(1-p)s}{r+s}\le(\frac{pr}{r}+\frac{(1-p)s}{s})=1\tag*{$\square$}$$
$3.$
$$\forall a,b,c,d>0,\frac{a}{c}+\frac{b}{d}>1\rightarrow\lim_{(x,y)→(0,0)}f(x, y) = 0.$$
Proof.
Let $a,b,c,d\in\mathbb(0,\infty)\cap{\mathbb{R}}$
Assume $\frac{a}{c}+\frac{b}{d}>1$
Show $\lim_{(x,y)→(0,0)} \frac{|x|^a|y|^b}{|x|^c+|y|^d}=0$
Let $p = \frac{ad}{c}-d+b$, that $\frac{a}{c}+\frac{b-p}{d}=1$, we can apply 2) on this
Implies $0\le\frac{|x|^a|y|^{b-p}}{|x|^c+|y|^d}\le1$
Have $$\lim_{(x,y)→(0,0)} \frac{|x|^a|y|^b}{|x|^c+|y|^d}$$
$$=\lim_{(x,y)→(0,0)} |y|^p\frac{|x|^a|y|^{b-p}}{|x|^c+|y|^d}=0\tag*{$\square$}$$
Best Answer
First, before doing anything, note that $f(0,y) = 0$ whenever $y \neq 0,$ so if the limit exists, it must equal $0.$ With this context, we can see that (1) is the contrapositive of $\Rightarrow$ and (3) is $\Leftarrow.$ As you see in the hints below, (2) is simply a lemma to prove (3).
Now for the hints (really, proof sketches):
For 1, set $t = |x|^c = |y|^d$ and let $t \to 0^+$.
For 2, set $w = |x|^c, z = |y|^d, t = \frac{a}{c}$ in Young's theorem.
For 3, factor out $|y|^p$ for a particular $p>0$ (chosen so that we may apply part 2 to the quotient) to show $$f(x,y) = |y|^p \cdot \text{ Some bounded function}$$