Show the function $x_1\sin(1/x_2)+x_2\sin(1/x_1)$ is continuous everywhere I am learning continuous function, please help me.
Show that the following function is continuous everywhere:
$\vec{F}(x_1,x_2)=x_1\sin{\left(\frac{1}{x_2}\right)}+x_2\sin{\left(\frac{1}{x_1}\right)}$ if $x_1x_2\neq 0$
and $\vec{F}(x_1,x_2)=0$ if $x_1x_2 = 0$
 A: Edit : Look at the comments for a cleaner (and in fact a correct) answer
I don't think this is actually continuous. For example consider the point $(1,0)$. We have $F(1,0)=0$.
Consider the line $(x,x-1)$, we have $\lim\limits_{x \to 1}(x,x-1)=(1,0)$.
Now $\lim\limits_{x \to 1} F(x,x-1) = \lim\limits_{x \to 1} x\sin(\frac 1 {x-1}) + (x-1)\sin(\frac 1 x) = \lim\limits_{x \to 1} x\sin(\frac 1 {x-1}) + \lim\limits_{x \to 1} (x-1)\sin(\frac 1 x)  $
With $\lim\limits_{x \to 1} (x-1)\sin(\frac 1 x) =0$ but $\lim\limits_{x \to 1} x\sin(\frac 1 {x-1})$ is undefined.
If the function was continuous we should have had $\lim\limits_{x \to 1} F(x,x-1) = F(\lim\limits_{x \to 1} (x,x-1)) = 0$
Hope I haven't done any mistakes :)
A: The function
$$F(x,y):=\cases{ x\sin{1\over y} + y\sin{1\over x}\quad &$\bigl((x,y)\ne(0,0)\bigr)$\cr
0 & $\bigl((x,y)=(0,0)\bigr)$\cr}$$
is continuous at all points $(x,y)$ with $xy\ne0$ and at $(0,0)$, and is discontinuous at all other points of ${\mathbb R}^2$.
Proof. 
(i) When $x_0 y_0\ne 0$ then in a full neighborhood of $(x_0,y_0)$ the upper alternative in the definition of $F(x,y)$ applies. Therefore $F$ is  continuous at $(x_0,y_0)$.
(ii) From $|F(x,y)|\leq |x|+|y|\leq \sqrt{2}\ \sqrt{x^2+y^2}$ it follows that $F$ is continuous at $(0,0)$. Indeed: Given an $\epsilon>0$ we have $|F(x,y)-0|\leq\epsilon$, as soon as $|(x,y)-(0,0)|=\sqrt{x^2+y^2}<\delta:={\epsilon\over\sqrt{2}}$.
(iii) Consider a point $(a,0)$ with $a> 0$. Then $F(a,0)=0$. On the other hand,
$$F(a,t)\geq a\sin{1\over t}- t\geq a\Bigl(\sin{1\over t}-{1\over2}\Bigr)\qquad\Bigl(0<t<{a\over2}\Bigr)\ .$$
Since there are arbitrary small $t$ in the given range with $\sin{1\over t}=1$ it follows that there is a sequence $(t_n)_{n\geq1}$ with $t_n\searrow 0$ $\ (n\to\infty)$ such that $F(a,t_n)\geq{a\over2}$ for all $n\geq1$. 
It follows that $F$ cannot be continuous at $(a,0)$.
A: The function is continuous since
$$ \left|x_1\sin{\left(\frac{1}{x_2}\right)}+x_2\sin{\left(\frac{1}{x_1}\right)}-0\right|\leq \left|x_1\sin{\left(\frac{1}{x_2}\right)}\right| + \left|x_1\sin{\left(\frac{1}{x_2}\right)}\right| $$
$$\leq |x_1|+|x_2| = \sqrt{x_1^2}+\sqrt{x_2^2} \leq \sqrt{{x_1}^2+{x_2}^2}+\sqrt{x_1^2+x_2^2} = 2 \sqrt{x_1^2+x_2^2} < \epsilon $$
$$ \implies \delta = \frac{\epsilon}{2}. $$
Note: We used the facts in the above derivations
$$ |\sin(t)|\leq 1 \,$$
$$ |x| = \sqrt{x^2}. $$
A: If this function were continuous everywhere (and we will see that it is not), then the one variable function $$f(t)=F(t,1)=\begin{cases}t\sin(1)+\sin\left(\frac1t\right)&t\neq0\\0&t=0\end{cases}$$ would be continuous. Since $t\sin(1)$ is continuous and equal to $0$ at $t=0$, this is equivalent to 
$$g(t)=\begin{cases}\sin\left(\frac1t\right)&t\neq0\\0&t=0\end{cases}$$ being continuous, but it is well known that this function is not continuous at $t=0$. Informally, as $t\to0$,  the outputs bounce back and forth more and more quickly between $-1$ and $1$.
