# 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 $x_1x_2 = 0$

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What do you know about a) the composition of continuous functions; b) the definition of continuity? –  Eckhard Dec 17 '12 at 14:21
@Coga81: Do you mean $F(x_1, x_2)$ instead of $F(u, v)$? Also, why do you have an arrow over $F$? –  Michael Albanese Dec 17 '12 at 14:34
Thank you very much, I am preparing basis for differential geometry. That why I wrote an arrow over F (vector function). It is written exactly from my book. –  Coga81 Dec 17 '12 at 18:10
It's unusual to call this function $\vec{F}$ rather than $F$, since there is only one output variable. It makes me wonder if there has not been a typo or transcription error somewhere. –  alex.jordan Aug 8 '14 at 0:18

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 :)

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The fact that your calculations do not work does not prove $F$ cannot be continuously defined. In particular, the fact that the individual limits do not exist does not prove that the sum does not (for example, consider $0=\lim 1/x - 1/x=\lim1/x-\lim1/x$). –  akkkk Dec 17 '12 at 14:56
right ! is it continuous though ? –  Zorba le Grec Dec 17 '12 at 14:59
I don't know, your reasoning sounds alright but you'll need to add a sentence somewhere - I'm not quite convinced. –  akkkk Dec 17 '12 at 14:59
Okay, so $\lim_{x\to1}F(x,x-1)-(x-1)\sin(1/x)$ should be defined by continuity of F and convergence of the term $(x-1)\sin(1/x)$ (convergence because $x-1\to0$ and $\sin$ is bounded). But this limit is your $x\sin(\frac{1}{x-1})$, which indeed does not converge since $\frac{1}{x-1}$ does not (add a few sentences here), ergo $F$ is not continuous. –  akkkk Dec 17 '12 at 15:04

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

$$|\sin(t)|\leq 1 \,$$

$$|x| = \sqrt{x^2}.$$

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This can only concern the behaviour of the function at the origin. The question asks about the behaviour of the function at every point. –  Did Dec 19 '12 at 16:41

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 $(a,0)$.

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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 $1$.

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