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I have to prove using (ε,δ)-definition of limit: $$ \lim_{(x,y) \to (7,2)} (x^2+y^2-xy) = 39 $$

I have already used the (ε,δ)-definition for simpler proofs in one variable and a few in two variables. However with this one I´ve been thinking all weekend without success.

The problem is to work with $$|x^2+y^2-xy-39|$$ and show that that is less than a formula involving δ, let´s call it g(δ), formula which I will later equal to ε in order to find the δs that guarantee: $$ ||(x,y)-(7,2)||<δ \implies |x^2+y^2-xy-39|<ε$$

I researched in the site and I haven´t seen a proof of a similar limit that could help me (namely, one of a 2-variable quadratic form). I tried different things, specifically, applying the triangle inequality in different steps. However, I ended up completing the square and landed this equality that gives me most hope (because constants cancel out):

$$|x^2+y^2-xy-39|=|(x-7)^2+(y-2)^2-(x-7)(y-2)+12(x-7)-3(y-2)|$$

If now I apply the triangle inequality I get:

$$ |x^2+y^2-xy-39|\le|x-7|^2+|y-2|^2+|x-7||y-2|+12|x-7|+3|y-2| $$

Here is where I get stuck, because if I apply $$ |x-7| \le ||(x-7,y-2)||$$ and $$ |y-2| \le ||(x-7,y-2)||$$ I arrive at $$ |x^2+y^2-xy-39|<3δ^2+15δ $$ which won´t be useful to get an explicit formula δ=δ(ε).

So, did I start well and then mess up or should I have done an entirely different thing?

Berni
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  • Why is $3\delta^2+15\delta-\varepsilon<0$ a problem? – MrYouMath May 02 '16 at 15:28
  • @MrYouMath That formula would be with = sign not < sign. Problem is that I need an explicit formula of δ in terms of ε and with that formula it´s not possible. – Berni May 02 '16 at 15:35
  • Note that for $\delta \le 1$, $\delta^2 \le \delta$, so for small enough $\delta$ you can bound $3\delta^2+15\delta& by $18\delta$. – Santiago Canez May 07 '16 at 20:57

2 Answers2

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$|(7+\delta)^2+(2+\delta)^2-(7+\delta)(2+\delta)-39| = |\delta(9+\delta)|=\delta(9+\delta)$, since $\delta>0$. Choose $\epsilon < \delta(9+\delta)$, and you're home free...

Doctor Dan
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Consider the solution of the system

$$3 \delta^2+15\delta-\epsilon=0$$

We will get the roots are

$$\delta =\frac{-15 \pm \sqrt{15^2+12\epsilon}}{6}$$

Since $\epsilon>0$, one of the root (say, $\delta_+$) is positive, and one of the root (say, $\delta_-$) is negative.

We are just interested in the region that $\delta>0$.

We can plot the curve $f(\delta)=3 \delta^2+15\delta-\epsilon$ and realized that $f(\delta)\leq 0$ for $\delta \in (0, \delta_+)$

Hence $\delta\leq \delta_+$ implies $f(\delta) \leq 0$ which implies $3 \delta^2+15\delta \leq \epsilon.$

Hence you can choose $$\delta=\frac{-15 + \sqrt{15^2+12\epsilon}}{6}$$

Credit to MrYouMath for raising the question.

Siong Thye Goh
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