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

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.

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 $a/c+b/d>1$.

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 $\lim\limits_{C\to 0^+} f(u_C) = 0$.