Help with Multivariable Delta-Epsilon Proof for $\lim_{(x,y)\to (1,2)}{ \left(2x^2 + y^2\right)} = 6$

Question

I'm in calculus 3 and feel like I understand how to verify single variable limits using delta-epsilon proofs pretty well. However, I a struggling more with verifying multivariable examples such as

$$\lim_{(x,y)\to (1,2)}{ \left(2x^2 + y^2\right)} = 6$$

I see lots of multivariable polynomial limit examples online but only of functions where $F(x,y)$ includes a product or quotient of variable expressions (I think I worded that correctly) and I understand most of those, but I really don't have any clue where to begin with one like this. Any push in the right direction would be appreciated!

Answer 1

$2x^2+y^2 - 6\\ 2x^2 - 4x + 2 + y^2 - 4y + 4 + 4x + 4y - 12\\ 2(x-1)^2 + (y-2)^2 + 4(x-1) + 4(y-2)$

Let our distance metric be the taxicab metric.

That is $d((x_0,y_0),(x_1,y_1)) = \max |x_1-x_0, y_1-y_0|$

if $d((x,y),(1,2)) \le \delta \le 1$ then $2(x-1)^2 + (y-2)^2 + 4(x-1) + 4(y-2) \le 11\delta$

Let $\delta = \max (1, \frac {\epsilon}{11})$

and it follows that $|f(x,y) - 6| < \epsilon$

An alternative would be to convert to a translated polar system.

Let $x = r\cos\theta + 1, y = r\sin\theta + 2,$ using the standard Euclidean metric, $d((x,y),(1,2)) = r$

or perhaps even better $x = \frac {r}{2} \cos\theta + 1, y = r\sin\theta + 2$

and $f(x,y)- 6=r(A\cos\theta + B\sin \theta)\le r\sqrt {A^2 + B^2} $

Answer 2

Let $x,y >0$, and $\epsilon >0$ be given.

Show that there is a $\delta >0$ such that

$(\star)$ $((x-1)^2 +(y-2)^2)^{1/2} \lt \delta$ implies

$f(x,y):= |2x^2-2 +y^2 -4| \lt \epsilon.$

$f(x,y) = |2(x^2-1) +y^4-4|=$

$|2(x-1)(x+1)+ (y-2)(y+2)|$

$\le 2(x+1)|x-1| +(y+2)|y-2|.$

Consider: $|x-1|\lt 1$, then

$0< x <2$, or $1< x+1< 3.$

Consider $ |y-2| \lt 1$, or

$1<y<3,$ or $3<y+2<5$.

Choose $\delta = \min (1, \epsilon/11)$

Then

$0 \le f(x,y) \le 6|x-1| + 5|y-2| \lt 11\delta \lt \epsilon.$

Note:

$(\star)$ implies

$ |x-1|\lt \delta$, and $|y-2| \lt \delta$.