Continuous function and limit $ f(x,y)=\left\{\begin{matrix} (x-y)\sin\frac{1}{x}\sin\frac{1}{y} & , xy\neq 0 \\ 0 &, x=y=0 \end{matrix}\right. $

Question

I have this function:$$ f(x,y)=\left\{\begin{matrix} (x-y)\sin\frac{1}{x}\sin\frac{1}{y} & , xy\neq 0 \\ 0 &, x=y=0 \end{matrix}\right. $$ a) Show that $ \lim_{x\rightarrow 0}[\lim_{y\rightarrow 0} f(x,y)] $ does not exist.

b) Show that $ f(x,y) $ is continuous at $ (0,0) $ .

For (a) I took $ \lim_{y\rightarrow 0} f(x,y) $ and I got that it's equal to $$ x\sin\frac{1}{x}\lim_{y\rightarrow 0}\sin\frac{1}{y} - \sin\frac{1}{x}\lim_{y\rightarrow 0}y \sin\frac{1}{y} $$, but $$ \lim_{y\rightarrow 0}\sin\frac{1}{y} $$ does not exist, so $ \lim_{y\rightarrow 0} f(x,y) $ does not exist and $ \lim_{x\rightarrow 0}[\lim_{y\rightarrow 0} f(x,y)] $ does not exist. Am I right?

For (b) I guess that I have to show that $$ \lim_{(x,y)\rightarrow (0,0)} f(x,y) = 0 $$ How can I do this ?

Answer 1

Use polar coordinates $x=r\cos\theta$ and $y=r\sin\theta$. The limit is the same as $r\to 0$. The original trigonometric functions are always between $-1$ and $1$, $|\cos\theta-\sin\theta|<2$, so your original expression is always less than $2r$ in absolute value.

Answer 2

For part (b), I think the easiest proof is just to note that $$|(x-y)\sin\frac{1}{x}\sin\frac{1}{y}| \leq |x-y|.$$

Thus, it's enough to prove that $$\lim_{(x,y) \to (0,0)}(x-y) = 0.$$

Answer 3

$sin{(1/x)sin(1/y)}$ is an interval and $x-y$ under the limit of $(x,y)\to{(0,0)}$ is an interval too. if a function in a close interval is continuous, its valued field is an interval, but its converse proposition may be not true. for example, your function is a counterexample. so you should separate your problem into two cases with $f^{'}(0,0)$ and $lim_{(x,y)\to{(0,0)}}f^{'}(x,y)$

the first limit exists but the second do not! for example rewrite your function to $F_{x}^{'}(x,y)=sin(1/x)sin(1/y)+(x-y)sin(1/y)cos(1/x)(-1/x^2)$

$F_{y}^{'}(x,y)=-sin(1/x)sin(1/y)+(x-y)sin(1/x)cos(1/y)(-1/y^2)$

divide both side with $sin(1/x)sin(1/y)$ to get

$1+(x-y)cos(1/x)(-\frac{1}{x^{2}sin(1/x)}）$

$-1+(x-y)cos(1/y)(-\frac{1}{y^{2}sin(1/y)）}$

since $f(x)=x^2{sin(1/x)}$ or equal to zero exists a limit at $f^{'}(0)$, but $lim_{x\to{0}}f^{'}(x)=2xsin(1/x)-cos(1/x)$ do not exists, this result give a certain conclusion to your problem. thank you!