show $\lim\limits_{(x,y)\to(0,0)}\frac{|x|^a|y|^b}{|x|^c+|y|^d}$ exists and equals $0$? suppose $f(x,y)=\frac{|x|^a|y|^b}{|x|^c+|y|^d}$ and $a,b,c,d >0$ how would you show that if $\frac{a}{c} + \frac{b}{d} >1$ then $\lim\limits_{(x,y)\to(0,0)}\frac{|x|^a|y|^b}{|x|^c+|y|^d}$ exists and equals $0$?
I've been trying to use the squeeze theorem and set up an inequality but im really struggling.
 A: Young's inequality states that for $x,y>0$ and $p,q>0$ with $\frac1p+\frac1q=1$ we have
$xy\le \frac{x^p}{p}+\frac{y^q}{q}$ so $|x|^a|y|^b\le \frac1p|x|^{ap}+\frac1q|y|^{bq}$. We can choose $p$ and $q$ such that $ap>c$ and $bq>d$:
Assume for a contradiction that for all $p,q$ with $\frac1p+\frac1q=1$ we have $ap\le c$ and $bq\le d$ then $\frac a c +\frac bd \le \frac1p+\frac 1q=1$. 
Now 
$$\frac{|x|^a|y|^b}{|x|^c+|y|^d} \le \frac{|x|^{ap}}{p |x|^c}+ \frac{|y|^{bq}}{q|y|^d}  \le |x|^{ap-c}+|y|^{bq-d},
$$
which tends to $0$ as $(x,y)\to (0,0)$.
A: Let consider


*

*$|x|=|u|^{1/c}$

*$|y|=|v|^{1/d}$


then we have that
$$\frac{|x|^a|y|^b}{|x|^c+|y|^d}=\frac{|u|^{a/c}|v|^{b/d}}{|u|+|v|}=r^{\frac a c+\frac b d-1}\cdot f(\theta)\to 0$$
since $f(\theta)=\frac{|\cos \theta|^{a/c}|\sin \theta|^{b/d}}{|\cos \theta|+|\sin \theta|}$ is bounded and for $\frac a c+\frac b d>1 \iff \frac a c+\frac b d-1>0$ we have that $r^{\frac a c+\frac b d-1}\to 0$.
A: The appropriate general technique is Lagrange multipliers. Because of the absolute value signs, we may ask for the largest value, with both $x,y > 0,$ of
$$ \frac{x^a y^b}{x^c + y^d} $$
The multiplier condition is that the ratio  $a x^{a-1} y^b  ::: c x^{c-1}$ is the same as  $b x^{a} y^{b-1}  ::: d y^{d-1}.$ Using cross-multiplication for fractions that are supposed to be equal gives
$$ adx^{a-1} y^{b+d-1} = bc x^{a+c-1} y^{b-1}.  $$ The first cancellation gives
$$ ad y^{b+d-1} = bc x^{c} y^{b-1}.  $$ The next gives
$$ ad y^{d} = bc x^{c} .  $$
That is, the largest value for fixed $x^c + y^d$ occurs when
$$ y = \lambda x^{c/d}   $$
and $\lambda$ is a positive constant.
$$ \frac{x^a y^b}{x^c + y^d} \leq \frac{\lambda_2 x^a x^{\frac{bc}{d}}}{x^c + \lambda_3 x^c} \leq \lambda_4 x^a x^{\frac{bc}{d}} x^{-c}$$
The resulting limit as $x$ goes to $0$ is also $0$ when
$$ a + \frac{bc}{d} - c > 0 \; , \; $$
$$ a + \frac{bc}{d} > c  \; , \; $$
or, as $c>0,$
$$  \frac{a}{c} + \frac{b}{d} > 1  \; . \; $$
