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.

• For the border case where $\frac{a}{c}+\frac{b}{d}=1$, see this question. – Arnaud D. Dec 3 '19 at 17:55

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

• Hi ! Can you give me some hints on how to check for the existence or non existence of limit in this case without the mod? – umm_what Aug 27 '20 at 16:49
• @juaps_: I’m not sure what you mean by without the mod. – Brian M. Scott Aug 27 '20 at 16:58
• I mean if it was just \frac{(x^a)(x^b)}{x^c + x^d} – umm_what Aug 27 '20 at 17:01
• @juaps_: Ah, okay; you mean without the absolute values. In that case the function may not even be defined: it depends on the exponents. – Brian M. Scott Aug 27 '20 at 17:06
• okay, thank you. Is it possible to derive some general relation between a,b,c,d in that case? – umm_what Aug 27 '20 at 17:12

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

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 $(u,v)\to 0$.