Help to understand limits in multivariable calculus

Question

Can someone show me some examples of limits by definition $\epsilon - \delta $ in multivariable calculus? I tried to understand in the books, but I´m quite confused right now.

Pd: If you know about a good book for this please tell me, I´m tryly confused.

Answer 1

Let $$f(x,y)= 3x+2y$$

Prove $$ lim_{(x,y)\to (1,5)} f(x,y)= 13 $$

Let $\epsilon >0 $ be given.

We need to fine a $\delta >0$ such that if $$\sqrt {(x-1)^2 + (y-5)^2 }< \delta $$

Then $$ | 3x+2y -13| < \epsilon $$

$$ | 3x+2y -13|= |(3x -3) +(2y -10)| \le 3|x-1| + 2|y-5|$$

Thus if $$|x-1|< \epsilon /6$$ and $$|y-5|< \epsilon /4$$

We get $$ | 3x+2y -13|< \epsilon $$

We know that we can make $$|x-1|< \epsilon /6$$ and $$|y-5|< \epsilon /4$$

Let $$ \delta = \epsilon /6 $$

If $$\sqrt {(x-1)^2 + (y-5)^2 }< \delta $$ Then we have $$|x-1|< \epsilon /6$$ and $$|y-5|< \epsilon /4$$

Thus $$ | 3x+2y -13| < \epsilon $$

And the proof is complete.

Answer 2

I don't have any books to recommend.

Instead, here an example.

$1$) Prove the following limit \begin{align} \lim_{(x, y)\rightarrow (0, 0)}\frac{xy\log(1+x^2+y^2)}{x^2+y^2} = 0 \end{align}

Pre-Proof/ Scratch Work: Observe we have that \begin{align} \left|\frac{xy\log(1+x^2+y^2)}{x^2+y^2}-0 \right| \le \frac{\frac{1}{2}(x^2+y^2)\log(1+x^2+y^2)}{x^2+y^2} = \frac{1}{2}\log(1+x^2+y^2). \end{align} Next, note that \begin{align} \log(1+t) \leq t \ \ \text{ whenever } \ \ t\geq 0 \end{align} which means \begin{align} \left| \frac{xy\log(1+x^2+y^2)}{x^2+y^2}\right|\leq \frac{1}{2}(x^2+y^2). \end{align} With this, we are ready to write down our $\varepsilon-\delta$ proof.

Proof: Fix $\varepsilon>0$. Choose $\delta = \sqrt{2\varepsilon}$. Then we see that \begin{align} \left| \frac{xy\log(1+x^2+y^2)}{x^2+y^2}\right| \leq \frac{1}{2}(x^2+y^2) < \frac{1}{2}\delta^2 = \varepsilon \end{align} whenever $\sqrt{x^2+y^2}<\delta$. $\square$

Here's another example (slightly more complicated):

$2$) Prove the following limit \begin{align} \lim_{(x, y)\rightarrow (1, 2)} \frac{xy}{1+x^2+y^2} =\frac{1}{3} \end{align}

Pre-Proof/ Scratch Work: Observe we have that \begin{align} \left|\frac{xy}{1+x^2+y^2}-\frac{1}{3} \right| =&\ \left|\frac{3xy-(1+x^2+y^2)}{3(1+x^2+y^2)}\right| = \frac{|3xy-1-x^2-y^2+2x-2x+4y-4y|}{3(1+x^2+y^2)}\\ =&\ \frac{|3(x-1)(y-2)-(x-1)^2-(y-2)^2+4(x-1)-(y-2)|}{3(1+x^2+y^2)}\\ \leq&\ \frac{1}{3}\left( 3|x-1||y-2|+|x-1|^2+|y-2|^2+4|x-1|+|y-2|\right)\\ \leq&\ \frac{1}{3}\left(|x-1|(3|y-2|+|x-1|+4)+|y-2|(|y-2|+1)\right)\\ \leq&\ \frac{1}{3}(|x-1|+|y-2|)(4|y-2|+4|x-1|+5)\\ \leq&\ \frac{\sqrt{2}}{3}\sqrt{(x-1)^2+(y-2)^2}\left(4\sqrt{2}\sqrt{(x-1)^2+(y-2)^2}+5 \right)\\ \leq&\ \frac{8}{3}\sqrt{(x-1)^2+(y-2)^2}\left(\sqrt{(x-1)^2+(y-2)^2}+2 \right). \end{align}

Proof: Fix $\varepsilon>0$ and choose $\delta = \sqrt{\frac{3}{8}}(\sqrt{\varepsilon+1}-1)$. Then we see that \begin{align} \left|\frac{xy}{1+x^2+y^2}-\frac{1}{3} \right| \leq \frac{8}{3}\delta(\delta+2)<\varepsilon \end{align} whenever $\sqrt{(x-1)^2+(y-2)^2}<\delta$. $\square$