Proving Multivariate Limit with Squeeze Theorem

Question

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.

Answer 1

Hint: Observe \begin{align} |x^2-6y^2| \leq |x+\sqrt{6}y||x-\sqrt{6}y| \leq (|x|+3|y|)|x-\sqrt{6}y|. \end{align}

Answer 2

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 $\max (|x|,|y|)\leq \sqrt {x^2+y^2}.$

Answer 3

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 $\Bbb R^n.$