# $\epsilon$-$\delta$ Proof of Limit in $\mathbb{R}^4$

Question: Determine $$\lim_{(a,b,c,d) \to (0, 0, 0, 0)} \frac{a^2d^2 -2abcd + b^2c^2}{a^2 + b^2 + c^2 + d^2}$$ and prove your result using the $$\epsilon$$-$$\delta$$ definition of a limit.

Attempt: I have completed the first part of the question using polar coordinates and found that the limit is $$0$$, but am having trouble proving my result.

Using $$z=(a,b,c,d)$$ and $$f(z)=\frac{a^2d^2 -2abcd + b^2c^2}{a^2 + b^2 + c^2 + d^2}$$.

RTP: $$\forall \epsilon>0, \exists \delta >0$$ s.t. ($$z \in \mathbb{R}^4$$ and $$0 < |z| < \delta$$) $$\Rightarrow$$ $$(|f(z)|< \epsilon)$$

However, this is at a level (particularly being in $$\mathbb{R}^4$$) I am not at yet.

Any help would be greatly appreciated.

Update: I've realised I'm not sure what $$0 < |z| < \delta$$ is, since $$|z| = |(a,b,c,d)|$$. Do I evaluate $$|(a,b,c,d)|$$ using the Euclidean norm?

• Does it help knowing that the numerator of $f(z)$ can be factorized? Namely,$$a^2d^2-2abcd+b^2c^2=(ad-bc)^2$$ Jun 2 '20 at 17:59
• @user170231 Would I use this for my $\delta$? I'm not sure of how much use this can be since the denominator cannot also be factorised to simplify $f(z)$. Jun 2 '20 at 18:04

you don't need an explicit coordinate system. Any $$\frac{(a,b,c,d)}{\sqrt{a^2 + b^2 + c^2 + d^2}}$$ is a unit vector; let us use different names, maybe $$(p,q,r,s)$$ such that $$p^2 + q^2 + r^2 + s^2 = 1.$$ Then your point $$(a,b,c,d) = t (p,q,r,s)$$ so that $$a^2 + b^2 + c^2 + d^2 = t^2$$ Next, $$(ad-bc)^2 = t^4 (pq-rs)^2$$ Now, how big can $$|pq-rs|$$ be? We know $$-1 \leq p,q,r,s \leq 1$$

• Why does it matter that $\frac{(a,b,c,d)}{\sqrt{a^2 + b^2 + c^2 + d^2}}$ is a unit vector? I haven't come across this when evaluating limits. Jun 2 '20 at 19:13
• @RubyPa you mentioned polar coordinates. There are also spherical coordinates in dimension 3, there is a radius called $\rho$ and two trig variables. We could do the same thing in dimension 4, a radius coefficient then 3 trig variables. Where did you get this problem? Jun 2 '20 at 19:19
• The problem is from my lecturer :) I used polar coordinates to compute it but our lecturer also wants us to learn how to prove our answers rigorously. Jun 2 '20 at 19:26
• The polar coordinates were useful to gain an intuition as to what the limit actually evaluates to and what I would need to set out to prove, but weren't 'technically' correct. Jun 2 '20 at 19:31
• @RubyPa that makes sense. The lecturer feels the class is fixated on methods that don't always work. Here is a problem where the $\delta, \epsilon$ part will be quite similar: $f(z)=\frac{a^2d^2 -2abcd + b^2c^2}{ \max(a^2 , b^2 , c^2 , d^2)}$ Jun 2 '20 at 19:31

The limit is $$0$$. To see this, start by writing $$\left| \frac{a^2d^2 -2abcd + b^2c^2}{a^2 + b^2 + c^2 + d^2}-0\right|=\frac{|a^2d^2 -2abcd + b^2c^2|}{a^2 + b^2 + c^2 + d^2}$$

$$\le \frac{a^2d^2}{a^2 + b^2 + c^2 + d^2}+\frac{2|abcd|}{a^2 + b^2 + c^2 + d^2}+\frac{b^2c^2}{a^2 + b^2 + c^2 + d^2}$$

$$=a^2\frac{d^2}{a^2 + b^2 + c^2 + d^2}+|cd|\frac{2|ab|}{a^2 + b^2 + c^2 + d^2}+b^2\frac{c^2}{a^2 + b^2 + c^2 + d^2}$$

$$\le a^2+ |cd|+b^2$$

Here, we have used the inequality $$2|ab|\le a^2+b^2$$ to bound the middle term. Then, note that the remaining fractions are all bounded above by $$1$$.

Applying $$|cd|\le c^2+d^2$$, we get that $$\left| \frac{a^2d^2 -2abcd + b^2c^2}{a^2 + b^2 + c^2 + d^2}-0\right|\le a^2+b^2+c^2+d^2$$

So, given $$\varepsilon>0$$, taking $$\delta=\sqrt{\varepsilon}$$ does the trick.

In more detail, suppose $$\varepsilon>0$$ is given. Write $$z=(a,b,c,d)$$ and put $$\delta:=\sqrt{\varepsilon}$$. Obviously, $$\delta>0$$. Now, if $$0<|z|<\delta$$, then

$$a^2+b^2+c^2+d^2=|z|^2<\delta^2=\varepsilon$$

Combining this with the previous inequalities gives the desired result.

• But if you have $|z|=|(a,b,c,d)|=\sqrt{a^2 + b^2 + c^2 + d^2}$, then wont $0 < |z| < \delta$ turn out to be $0 < \sqrt{a^2 + b^2 + c^2 + d^2} < \delta$ meaning that $a^2 + b^2 + c^2 + d^2 \not\le \epsilon$? Jun 2 '20 at 23:07
• @RubyPa Oh, you're right, my apologies. Still, not all is lost, and you can take $\delta=\sqrt{\varepsilon}$. Then $|z|<\delta$ implies $a^2+b^2+c^2+d^2<\varepsilon$. Jun 2 '20 at 23:09
• Thanks for your help :) Jun 2 '20 at 23:18