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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?

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    $\begingroup$ Does it help knowing that the numerator of $f(z)$ can be factorized? Namely,$$a^2d^2-2abcd+b^2c^2=(ad-bc)^2$$ $\endgroup$
    – user170231
    Jun 2 '20 at 17:59
  • $\begingroup$ @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 $f(z)$. $\endgroup$
    – Ruby Pa
    Jun 2 '20 at 18:04
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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.

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  • $\begingroup$ 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 $a^2 + b^2 + c^2 + d^2 \not\le \epsilon$? $\endgroup$
    – Ruby Pa
    Jun 2 '20 at 23:07
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    $\begingroup$ @RubyPa Oh, you're right, my apologies. Still, not all is lost, and you can take $\delta=\sqrt{\varepsilon}$. Then $|z|<\delta$ implies $a^2+b^2+c^2+d^2<\varepsilon$. $\endgroup$
    – Reveillark
    Jun 2 '20 at 23:09
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    $\begingroup$ Thanks for your help :) $\endgroup$
    – Ruby Pa
    Jun 2 '20 at 23:18
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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 $-1 \leq p,q,r,s \leq 1$

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  • $\begingroup$ 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. $\endgroup$
    – Ruby Pa
    Jun 2 '20 at 19:13
  • $\begingroup$ @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? $\endgroup$
    – Will Jagy
    Jun 2 '20 at 19:19
  • $\begingroup$ 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. $\endgroup$
    – Ruby Pa
    Jun 2 '20 at 19:26
  • $\begingroup$ 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. $\endgroup$
    – Ruby Pa
    Jun 2 '20 at 19:31
  • $\begingroup$ @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: $f(z)=\frac{a^2d^2 -2abcd + b^2c^2}{ \max(a^2 , b^2 , c^2 , d^2)}$ $\endgroup$
    – Will Jagy
    Jun 2 '20 at 19:31

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