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Given that $\frac{a}{c} + \frac{b}{d} > 1$, I am attempting to show that $$\lim_{(x,y)\rightarrow(0,0)} \frac{|x|^{a}|y|^{b}}{|x|^{c} + |y|^{d}} = 0$$

My attempt at a solution:

Let's assume the limit is in fact zero, and prove it with the Squeeze Theorem. We have that \begin{align*} \Bigg|\frac{|x|^{a}|y|^{b}}{|x|^{c} + |y|^{d}}\Bigg| &= \frac{|x|^{a}|y|^{b}}{|x|^{c} + |y|^{d}} \\ &\leq \frac{1}{2}|x|^{a-\frac{c}{2}}|y|^{b-\frac{d}{2}} && \text{using the fact that } a^{2} + b^{2} > 2ab \end{align*}

But I'm not quite sure how to proceed from here to apply the inequality given and find a limiting function, if I am even on the right track with the inequality I applied. This isn't homework ( a practice question I found ), any help would be greatly appreciated.

Fabrosi
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miradulo
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1 Answers1

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I think I've got it! I was on the wrong track after all, the key is to turn it into a single variable limit. If there's any issues with this result please do let me know!

Letting $|x| = q^{\frac{1}{c}}$ and $|y| = q^{\frac{1}{d}}$, we have \begin{align*} \frac{|x|^{a}|y|^{b}}{|x|^{c} + |y|^{d}} &= \frac{q^\frac{a}{c}q^\frac{b}{d}}{2q} \\ &= \frac{1}{2}q^{\frac{a}{c} + \frac{b}{d} -1} \end{align*} Then as \begin{align*} \lim_{(x,y)\rightarrow(0,0)}\frac{1}{2} q^{\frac{a}{c} + \frac{b}{d} -1} &= \frac{1}{2}\lim_{q\rightarrow0} q^{\frac{a}{c} + \frac{b}{d} -1} \\ &= \frac{1}{2}\lim_{q\rightarrow0} q^{k} &&\text{where } k>0 \\ &= 0 \end{align*} by the Squeeze Theorem we have that shown that the limit is in fact $0$.

miradulo
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