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?
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.