<s>[INST] Help with definition of derivative My textbook says the definition is this:
$$\lim\limits_{\textbf{x} \rightarrow \textbf{x}_0} \frac{\|f(\textbf{x}) - f(\textbf{x}_0) - \textbf{T}(\textbf{x} - \textbf{x}_0)\|}{\|\textbf{x} - \textbf{x}_0\|} = 0$$
I am trying to learn how to use it on, say for example, the function $f(x,y) = xy$ at the point $(1, 1)$.
So we have:
$$\lim\limits_{(x,y) \rightarrow (1,1)} \frac{\|xy - 1 - \textbf{T}(\textbf{x} - \textbf{x}_0)\|}{\|(x,y) - (1,1)\|} = 0$$
$$\lim\limits_{(x,y) \rightarrow (1,1)} \frac{\|xy - 1 - \textbf{T}(\textbf{x} - \textbf{x}_0)\|}{\sqrt{(x - 1)^2 + (y - 1)^2}} = 0$$
$$\lim\limits_{(x,y) \rightarrow (1,1)} \frac{\|xy - 1 - [y(x - 1) + x(y - 1)]\|}{\sqrt{(x - 1)^2 + (y - 1)^2}} = 0$$
$$\lim\limits_{(x,y) \rightarrow (1,1)} \frac{\|xy - 1 - xy + y - xy + x\|}{\sqrt{(x - 1)^2 + (y - 1)^2}} = 0$$
$$\lim\limits_{(x,y) \rightarrow (1,1)} \frac{\|- 1 + y - xy + x\|}{\sqrt{(x - 1)^2 + (y - 1)^2}} = 0$$
But if I plug in $(1,1)$, I get an indeterminate form $\frac{0}{0}$. Is everything correct so far?
 [/INST] Notice taking limit $x \neq 1$, and $y \neq 1$, but they get very close to $(1,1)$, and this is enough to get through the problem...
So starting at your last line: $|| - 1 + y - xy + x|| = |x - 1|\cdot |y - 1|$, so:
$0 \leq \dfrac{||-1 + y - xy + x||}{\sqrt{(x - 1)^2 + (y - 1)^2}} = \dfrac{|x - 1|\cdot |y - 1|}{\sqrt{(x - 1)^2 + (y - 1)^2}} \leq |x - 1|$. So:
$|x - 1| \to 0$ when $x \to 1$, and by squeeze theorem the middle term $\to 0$.
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