multivariable continuous

Question

I wanna show that for $\alpha+\beta-2\gamma>0$:

$\lim_{(x,y)\rightarrow(0,0)}\frac{\vert x\vert^{\alpha}\cdot\vert y\vert^{\beta}}{(x^{2}+y^{2})^{\gamma}}=0$

I thought about proving it via Sandwhich theory but I have no idea how to simplify this expression.

Thanks in advance!

Answer 1

I thought it might be instructive to follow the OP's idea to use the squeeze (sandwich) theorem and present a way forward accordingly. To that end, we proceed.

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From the AM-GM inequality, we note that $x^2+y^2 \ge 2|xy|$.

Applying this to $\frac{|x|^\alpha|y|^\beta}{(x^2+y^2)^\gamma}$, we find that

$$0\le \frac{|x|^\alpha|y|^\beta}{(x^2+y^2)^\gamma}\le \frac{|x|^\alpha|y|^\beta}{2^\gamma |x|^\gamma|y|^\gamma}= 2^{-\gamma}\max(|x|^{\alpha+\beta-2\gamma},|y|^{\alpha+\beta-2\gamma})$$

And now, applying the squeeze theorem, we find the coveted limit is $0$ when $\alpha +\beta-2\gamma>0$.

Answer 2

Put $x=r\cos\theta, y=r\sin\theta$, then your expression becomes $r^{\alpha+\beta-2\gamma}|\sin\theta|^{\beta}|\cos\theta|^{\alpha}$, which tends to zero if $\alpha+\beta-2\gamma>0$.

Answer 3

Hint. One may use polar coordinates, $$ x=r \cos \theta,\qquad y=r \sin \theta, $$ giving $$ \frac{\vert x\vert^{\alpha}\cdot\vert y\vert^{\beta}}{(x^{2}+y^{2})^{\gamma}}=r^{\alpha+\beta-2\gamma}\left|\cos \theta \right|^\alpha\left|\sin \theta \right|^\beta. $$ Can you take it from here?