Multivariable limit proof: $\lim\limits_{(x,y)\rightarrow (0,0)}\frac{\left|x\right|^a\left|y\right|^b}{\left|x\right|^c + \left|y\right|^d} = 0$ 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.
 A: There is an easy solution. Basically just change the numerator to:
$(|x|^c)^{a/c} (|y|^d)^{b/d}$ 
and use the inequalities:
$|x|^c \leq |x|^c + |y|^d$ and $|y|^d \leq |x|^c + |y|^d$
then cancel out and use squeeze theorem.
Done.
A: Since we're trying to prove that a limit of a nonnegative quantity equals zero, all we need to do is find an upper bound that tends to zero (the squeeze theorem). The simplest upper bound for a sum in a denominator $1/(f+g)$, if $f$ and $g$ are both nonnegative, is that it's at most $1/f$ and also at most $1/g$. Furthermore, when $f\ge g$ then the upper bound $1/(f+g) \le 1/f$ is off by at most a factor of 2 (and similarly $1/(f+g) \le 1/g$ when $g\ge f$), so it's not likely to make things much worse.
So my hint is to find, in this way, two upper bounds for the function in question, one that holds when $|x|^c \ge |y|^d$ and one that holds when $|x|^c \le |y|^d$. Both upper bounds will turn out to tend to zero when $(x,y)$ tends to the origin, thanks to the assumption $a/c+b/d>1$.
A: Let $u = \vert x\vert^c$ and $v = \vert y\vert^d$; then you’re essentially interested in $\lim\limits_{(u,v)\to (0,0)}\dfrac{u^{a/c}v^{b/d}}{u+v}$. For convenience let $\alpha = \dfrac{a}{c}$ and $\beta = \dfrac{b}{d}$, and consider $\lim\limits_{(u,v)\to (0,0)}\dfrac{u^\alpha v^\beta}{u+v}$, where $\alpha + \beta > 1$.
Notice that no matter how $(u,v)$ approaches the origin, $u+v$ must approach $0$. Conversely, if $u+v \to 0$, then $(u,v)\to (0,0)$. This suggests looking at the worst-case (= largest) value of $\dfrac{u^\alpha v^\beta}{u+v}$ for a fixed value of $u+v$.
For the moment, then, let’s fix some positive real number $C$ and consider only values of $\dfrac{u^\alpha v^\beta}{u+v}$ for which $u+v=C$. We can then rewrite $\dfrac{u^\alpha v^\beta}{u+v}$ as $f(u) = \dfrac1C u^\alpha (C-u)^\beta$. Then $$\begin{align*}
f'(u) &= \frac1C\left(\alpha u^{\alpha-1}(C-u)^\beta - \beta u^\alpha (C-u)^{\beta-1}\right)\\
&= \frac{u^{\alpha-1}(C-u)^{\beta-1}}{C}(\alpha(C-u)-\beta u)\\
&= \frac{u^{\alpha-1}(C-u)^{\beta-1}}{C}(\alpha C - (\alpha+\beta)u).
\end{align*}$$
Now $u$ and $v$ are both positive, so $0 < u < C$, and $f'(u)=0$ only at $u_C = \dfrac{\alpha C}{\alpha+\beta}$. It’s also clear that $f'(u) > 0$ when $0 < u < u_C$ and $f'(u) < 0$ when $u_C < u < C$, so $f(u)$ has a relative maximum at $u_C$. At this maximum we have
$$\begin{align*}
f(u_C) &= \frac1C u_C^\alpha(C - u_C)^\beta\\
&= \frac1C \cdot \frac{\alpha^\alpha C^\alpha}{(\alpha+\beta)^\alpha} \cdot \frac{\beta^\beta C^\beta}{(\alpha+\beta)^\beta}\\
&= \frac{\alpha^\alpha \beta^\beta C^{\alpha+\beta}}{C(\alpha+\beta)^{\alpha+\beta}}\\
&= \frac{\alpha^\alpha \beta^\beta}{(\alpha+\beta)^{\alpha+\beta}} C^{\alpha+\beta-1}
\end{align*},$$
where $\dfrac{\alpha^\alpha \beta^\beta}{(\alpha+\beta)^{\alpha+\beta}}$ is a constant independent of $C$. Here, finally, is where we use the fact that $\alpha+\beta > 1$: $\alpha+\beta - 1 > 0$, so $\lim\limits_{C\to 0^+} C^{\alpha+\beta-1} = 0$, and hence $\lim\limits_{C\to 0^+} f(u_C) = 0$.
A: 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 $(u,v)\to 0$.

