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Examine the continuity and discontinuity of the following function at $(0,0)$

$$f(x,y)=\begin{cases}{x^3\cos({1\over y})+y^3\sin({1\over x})\over {x^2+y^2}} & x\neq 0\neq y\\ 0 & \text{otherwise}\end{cases}$$

I tried to prove continuity using the definition of limits

$$|f(x,y)-f(0,0)|=|{x^3\cos({1\over y})\over x^2+y^2}|+|{y^3\cos({1\over x})\over x^2+y^2}|$$

$$\le|{x^3+y^3\over x^2+y^2}|$$

what do i do next ? Need some hint on proving differentiability as well .

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How do we define $f$ where $x=0\neq y$ (or vice versa)? – Jonathan Y. Oct 2 '13 at 20:57
Naturally, since there are points matching this description in any neighborhood of zero, the way the function is defined there is essential to the solution of the exercise. As the answers imply, the only way to get a limit at the origin is to have the function tend to zero along the axes as well (and whatever we might do won't help with differentiability). – Jonathan Y. Oct 2 '13 at 22:12
Your change was superficial: it didn't change the definition of the function. I proposed a different change that I suspect was in the original question; please review it. If I'm right, then $f_x$, $f_y$ do exist and are zero at the origin, but $\frac{|f(x,y)|}{\sqrt{x^2+y^2}}$ has no limit as $(x,y)\to (0,0)$. – Jonathan Y. Oct 3 '13 at 7:14
Oh, I see now @njguliyev suggested (and addressed) that change 10 hours ago. – Jonathan Y. Oct 3 '13 at 7:16
It's the definition of differentiability: $\frac{|f(x,y) - f(0,0) - \langle\nabla f(0,0),(x,y)\rangle|}{\|(x,y)\|}$ – Jonathan Y. Oct 3 '13 at 7:21
up vote 2 down vote accepted

Hint: $$\left|\frac{x^3}{x^2+y^2}\right| = |x|\left|\frac{x^2}{x^2+y^2}\right|\le |x|.$$

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how did you establish the last inequality ? – Aman Mittal Oct 2 '13 at 20:41
$x^2+y^2 \ge x^2$. – njguliyev Oct 2 '13 at 20:42
oh ! Nice, and some hint of differentiability please ! – Aman Mittal Oct 2 '13 at 20:44
By the way, you didn't define your function at points of the form $(x,0)$ and $(0,y)$, but I assume by continuity that $f=0$ at those points. Then $f_x(0,0) = f_y(0,0)=0$. Hence differentiability is equivalent to $f(x,y)=o(\sqrt{x^2+y^2})$. That is you have to check whether $\frac{f(x,y)}{\sqrt{x^2+y^2}}\to 0$ or not. – njguliyev Oct 2 '13 at 21:05

Hint Try polar coordinates: $(x,y)=(r\cos\theta,r\sin\theta)$. Your bound at the RHS decomposes into a radial component (limiting to zero as $r\downarrow0$) and a bounded angular component.

Differentiability: Jonathan Y. correctly pointed out that neither of the axes (except for the origin) belongs to your domain of definition, so you can't use nonexistence of partial derivatives. We still need to show that one cannot find a & b such that $f(x,y)-f(0,0) = a x + b y + \varepsilon_x x + \varepsilon_y y$, with $\varepsilon_x\to0$ and $\varepsilon_y\to0$ as $(x,y)\to(0,0)$. There's nothing special about partial derivatives: if the function is differentiable, then all admissible directional derivatives exist (i.e., in all directions that locally lie within the domain of definition). Use, then, $f(x,x)= x \frac{\cos(1/x)+\sin(1/x)}{2}$. If the function was differentiable, that limit as $x\to0$ would equal $a+b$ - here, it does not exist, which is a contradiction.

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should i use polars on the last inequality i get ? – Aman Mittal Oct 2 '13 at 20:35
@AmanMittal - yes! Re: differentiability, what have you tried out? – automaton 3 Oct 2 '13 at 20:37
i get something like this $|r||\cos^3 \theta+\sin^3 \theta|$ But i still dont see how that leads to a limit – Aman Mittal Oct 2 '13 at 20:41
well, for diff i tried using the definition - $f(0+h,0+k)-f(0,0)=hA+kB+h\phi (h,k)+k \theta (h,k)$ but didnt reach anywhere :) – Aman Mittal Oct 2 '13 at 20:42
The angular component is trivially bounded by $2$. As for $r$, it approaches zero because $(x,y)=(0,0)$ corresponds to $r=0$. – automaton 3 Oct 2 '13 at 20:43

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