Use the epsilon-delta definition of limits for a multivariable function Use the epsilon-delta definition of limits to show that $\lim \limits_{(x,y) \to (1,2)}(2x^{2}+y^{2}) = 6$
I tried putting it in polar coordinates and got to $\lim \limits_{r \to \sqrt
6}r^2 = 6$
From there I got $|r-\sqrt{6}| < \delta$, how do I get that to the form $|r^2 - 6| < \epsilon$? Or is there a simpler way to do the problem?
 A: You didn't specify that to use polar coordinates we need firstly to set


*

*$u^2=2x^2$

*$v^2=y^2$


then 
$$\lim \limits_{(x,y) \to (1,2)}(2x^{2}+y^{2}) = 6\iff \lim \limits_{(u,v) \to (\sqrt 2,2)}(u^{2}+v^{2}) = 6\iff\lim \limits_{r \to \sqrt 6}r^2 = 6$$
From here you can easily proceed by the $\epsilon-\delta$ definition of limit, that is
$$|r^2 - 6| < \epsilon\iff \sqrt{6-\epsilon}<r<\sqrt{6+\epsilon}\iff \sqrt{6-\epsilon}-\sqrt 6<r-\sqrt6<\sqrt{6+\epsilon}-\sqrt 6$$
then the optimal bound for $\delta$ is
$$|r-\sqrt 6|<min\{\sqrt 6-\sqrt{6-\epsilon},\sqrt{6+\epsilon}-\sqrt 6\}=\sqrt{6+\epsilon}-\sqrt 6=\delta$$
Or as an alternative note that if we assume $$|r-\sqrt 6|<1 \implies r\in(-1+\sqrt 6,1+\sqrt 6)\implies r+\sqrt 6<1+2\sqrt 6$$
we have
$$|r^2-6|=(r+\sqrt 6)|r-\sqrt 6|<(1+2\sqrt 6)|r-\sqrt 6|$$
and thus if
$$|r-\sqrt 6|<\frac{\epsilon}{1+2\sqrt 6}\implies|r^2-6|<\epsilon$$
and $$\delta =\min\left\{1, \frac{\epsilon}{1+2\sqrt 6}\right\}$$
A: \begin{align*}
|2x^{2}+y^{2}-6|&=|2x^{2}+y^{2}-2(1)^{2}-2^{2}|\\
&=2|x^{2}-1|+|y^{2}-2^{2}|\\
&\leq 2|x-1||x+1|+|y-2||y+2|\\
&<2\cdot 3|x-1|+5|y-2|\\
&\leq 30(|x-1|+|y-2|)\\
&\leq 60\sqrt{(x-1)^{2}+(y-2)^{2}}\\
&<60\cdot\epsilon/60\\
&=\epsilon,
\end{align*}
where $\sqrt{(x-1)^{2}+(y-2)^{2}}<\min\{1,\epsilon/60\}$.
