TITLE: Multivariable limit proof: $\lim\limits_{(x,y)\rightarrow (0,0)}\frac{\left|x\right|^a\left|y\right|^b}{\left|x\right|^c + \left|y\right|^d} = 0$
QUESTION [9 upvotes]: I have some trouble with this: Show that if $a, b \ge 0$, $c, d > 0$ and $\frac{a}{c} + \frac{b}{d} > 1$ then 
$$\lim_{(x,y)\rightarrow (0,0)}\frac{\left|x\right|^a\left|y\right|^b}{\left|x\right|^c + \left|y\right|^d} = 0$$
we were not really shown how to evaluate multivariable limits other than trying different paths, which wouldn't work when the limit exists, or by applying the squeeze theorem which is my initial plan. Unfortunately I cannot find a proper bound for the denominator and so I am quite lost right now. Any help would be appreciated.

REPLY [3 votes]: Using the same simplification that Brian M. Scott, we need bound
$u^\alpha v^\beta$ for $u\ge0$, $v\ge 0$, $\alpha + \beta > 1$. By symmetry, we have two cases:

$0\le\beta\le 1$. Applying the weighted AM-GM inequality with weights $(1 - \beta),\beta$:
$$
u^\alpha v^\beta = u^{\alpha + \beta - 1}u^{1 - \beta}v^\beta\le
u^{\alpha + \beta - 1}((1 - \beta)u + \beta v)\le
u^{\alpha + \beta - 1}(u + v).
$$
$\beta > 1$. We can do simply:
$$
u^\alpha v^\beta = u^\alpha v^{(\beta - 1)}v\le
u^\alpha v^{(\beta - 1)}(u + v).
$$
And in both cases $u^\alpha v^\beta/(u + v)\to 0$ when $(u,v)\to 0$.