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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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 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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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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