# Proving Multivariate Limit with Squeeze Theorem

How to prove that

\begin{align} \lim_{(x,y) \rightarrow (0,0)} \frac{x^2-6y^2}{|x| + 3|y|} = 0\ \end{align}

using the Squeeze Theorem? I can work the limit down to $\frac{|x^2 -6y^2|}{|x+y|}$ but can't find a ball $B(x)$ to make the Squeeze Theorem work.

Let $z=\max (|x|,|y|).$ We have $|x^2-6y^2|\leq |x|^2+6|y|^2\leq 7z^2.$ We have $|x|+3|y|\geq z.$ So when x,y are not both $0$ we have $$\left| \frac {x^2-6y^2}{|x|+3|y|}\right| \leq \frac {7z^2}{z}=7z=7\max (|x|,|y|)\leq 7\sqrt {x^2+y^2}.$$ Note: We have $|x|\leq \sqrt {x^2+y^2}$ and $|y|\leq \sqrt {x^2+y^2},$ so $\max (|x|,|y|)\leq \sqrt {x^2+y^2}.$
• Note that to prove convergence to zero, we don't necessarily need a precise upper bound, Just one that works, like $7z.$ Oct 18, 2016 at 3:48
Alternatively: $$0\le\left|\frac{x^2-6y^2}{|x|+3|y|}\right|=\frac{|x^2-6y^2|}{|x|+3|y|}\le\frac{x^2+6y^2}{|x|+|y|}\le6\frac{x^2+y^2}{|x|+|y|}\le6\frac{(|x|+|y|)^2}{|x|+|y|}=6(|x|+|y|)\to 0$$ by the continuity of norms on $$\Bbb R^n.$$