I don't understand  $(x \notin A \setminus B)$  Why is it not $$x \notin A \land x\in B$$ instead of $$x \notin A \lor x\in B$$
Given two sets$$A=\{1,3,5\}\quad,\quad B=\{1,3,8\}.$$ Then I compute the $A\setminus B=\{5\}.$   But my book$^\dagger$said: ... . Also, observe that $x\notin A\setminus B$ does not mean that $x\notin A\lor x\in B$. Why?  I don't know how to explain the Why, please give me some example.    $^\dagger$, second edition, page $4$.  EDIT: My intuition tells me that it does mean that. I guess there is some special case about null-set or about the restriction of the universal set.
Let $f: \mathbb R^2 \to \mathbb R$ defined as   $f(x,y)=(x^2+y^2)\sin(\dfrac{1}{\sqrt{x^2+y^2}})$ if $(x,y) \neq (0,0)$, $f(x,y)=(0,0)$ if $(x,y)=(0,0)$  Prove that $f$ is differentiable at $(0,0)$ but that the partial derivatives of $f$ are not continuous at the origin.  The attempt at a solution:  I had some problems at the first part of the exercise:  I've calculated $\dfrac{\partial f}{\partial x}(0,0)$ and $\dfrac{\partial f}{\partial y}(0,0)$, both exist and are equal to $1$.  So, to prove $f$ is differentiable at the origin, I have to show that   $\lim_{(x,y) \to (0,0)} \dfrac{|f(x,y)-(f(0,0)+&lt;\nabla f(0,0),(x,y)&gt;)|}{\|(x,y)\|}=0$ But $\lim_{(x,y) \to (0,0)} \dfrac{|f(x,y)-(f(0,0)+&lt;\nabla f(0,0),(x,y)&gt;)|}{\|(x,y)\|}$ equals  $\lim_{(x,y) \to (0,0)} \dfrac{|(x^2+y^2)\sin(\dfrac{1}{\sqrt{x^2+y^2}})-(x+y)|}{\|(x,y)\|}$. I don't know how to show this limit equals $0$.  For the second part (I've only analyzed $\dfrac{\partial f}{\partial x}$ since both cases are the same), what I did was try to show that $\lim_{(x,y) \to (0,0)} \dfrac{\partial f}{\partial x}(x,y) \neq \dfrac{\partial f}{\partial x}(0,0)$   I thought of considering the curve $\phi(t)=(t,0)$, so   $\lim_{t \to 0} (f\circ \phi)(t)=\lim_{t \to 0} 2t\sin(|t|^{-1})-\dfrac{t\cos(|t|^{-1})}{|t|^{-1}}$. But $\lim_{t \to 0^{+}} \dfrac{t\cos(|t|^{-1})}{|t|^{-1}}=\lim_{t \to 0^{+}} \cos(\dfrac{1}{t})$ and this limit doesn't exist, so the original limit doesn't exist. This shows the partial derivative with respect to $x$ at the point $(0,0)$ is not continuous.   I would appreciate if someone could help me to finish the first part and tell me if what I did in the second part of the problem is correct. In vanilla Gmail, clicking on a conversation alters the fragment ID in the URL, allowing you to link back to that conversation.  (It opens in its own Gmail interface.)  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