Proving limits with $\epsilon$ - $\delta$ -definition for 2 variable functions

Question

I know there are a lot of questions considering how to prove limits for two variable functions. But I didn't find any examples about what I am going to ask now.

So how do I proceed in situations when my function is in form like $f(x,y)=x^2+xy$ and I have to prove that $\lim f(x)=6$ when $(x,y) \to (2,1)$?

I have a lot of examples when functions also have a denominator (like this one here: includes also the definition of a limit) but how can I get from here $|x^2 + xy -4| < \epsilon$ to there $||(x,y)-(2,1)||=\sqrt{(x-2)^2 +(y-1)^2} < \delta$?

Thanks a lot!

Answer 1

I'm assuming you mean limit 6 instead of 4.

When $\left\| (x,y)-(2,1) \right\| < \delta$, you surely have $|x-2| < \delta$ and $|y-1| < \delta$. Rewrite:

$$\begin{align} \color{blue}{x^2}+\color{red}{xy}-6 & = \color{blue}{(x-2)^2+4x-4}+\color{red}{(x-2)(y-1)+x+2y-2}-6 \\ & = (x-2)^2+(x-2)(y-1)+5(x-2)+2(y-1) \end{align}$$

You then have, choosing $\delta < 1$: $$\begin{align} \left| x^2+xy-6\right| & \le |x-2|^2+|x-2||y-1|+5|x-2|+2|y-1| \\ & \le 2\delta^2+5\delta+2\delta \\ & \le 9\delta \end{align}$$

You want this under $\varepsilon$. Can you finish from here?

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Side note:

but how can I get from here $|x^2 + xy -4| < \epsilon$ to there $||(x,y)-(2,1)||=\sqrt{(x-2)^2 +(y-1)^2} < \delta$?

You have it the wrong way, you need to get from "$\ldots < \delta$" to "$\ldots < \varepsilon$", not the other way around. Take a good look at the definition ("if $\ldots < \delta$, then $\ldots < \varepsilon$").

Answer 2

It's the same as to prove that $|x-a|<\delta $ and $|y-b| <\delta$. Did you mean to prove that $\lim_{(x,y)\to (2,1)} f(x,y) =6?$ In that case you can just substitute your $x$ and $y$ by the given $(a,b)$ and you will get $2^2 +2*1=6$. If there is a denominator then it's another story to lose that denominator or to prove that the limit doesn't exist.