# Examining continuity and differentiability of function

Let $$F(x,y) = \begin{cases} \frac{x³\cos(1/y) + y³\cos(1/x)}{(x²+y²)} & \text{for x,y ≠ (0,0)} \\ 0 & \text{otherwise}\end{cases}$$

How to examine the continuity at $(0,0)$

• $x,y\ne (0,0)$ doesn't make sense. – zhw. Dec 20 '16 at 20:26
• I'm sure he means $(x,y) \neq (0,0)$. – David Dec 20 '16 at 20:27
• What is $F(1,0)$ then? – zhw. Dec 20 '16 at 20:28

How to examine continuity at $(0,0)$?

A function $f$ is continuous at $(a,b)$ if $$\lim_{(x,y)\rightarrow (a,b)}f(x,y)=f(a,b)$$ just compute the limit.

Hint for limit: Convert to polar coordinates and express limit as $r*(\text{a bunch of bounded stuff})$

• can you help me with calculating the limit? – user383053 Dec 20 '16 at 19:50
• @YashwanthKasturi I added a hint. Let me know if this helps – qbert Dec 20 '16 at 19:52
• No buddy... im getting stuck in between... i took x=r cosa; y = r sina.. but i am not able to eliminate indeterminate form from limit – user383053 Dec 20 '16 at 20:09
• @qbert Actually , I had a related question. Is it possible without graph to check whether a given function is continuous in its entire domain . There are infinite points and checking continuity at end points can't give the answer – Shashaank Dec 20 '16 at 20:11
• Sure. If you can show continuity for any arbitrary point in the domain, then you have shown continuity of the function everywhere. Try with $f(x)=x^2$. – qbert Dec 20 '16 at 20:14

Let $x = r \cos\theta$ and $y = r \sin \theta$. Of course, $r = \sqrt{x^2+y^2}$. Rewrite $F(x,y)$ as $F(r,\theta)$, that is for $r>0$:

$F(r,\theta) = \frac{r^3\cos^3\theta\cos{(1/r\sin \theta)+r^3 \sin^3\theta\cos(1/r\cos \theta)}}{r} \\= r^2\cos^3\theta\cos{(1/r\sin \theta)+r^2 \sin^3\theta\cos(1/r\cos \theta)}.$

Then, consider a delta-epsilon proof noting that

$\mid F(r,\theta)\mid \;\leq \; \mid r^2\cos^3\theta\cos{(1/r\sin \theta)} \mid + \mid r^2 \sin^3\theta\cos(1/r\cos \theta) \mid \; \leq \;2r^2$

(where the first inequality comes from the triangle inequality).

• But what about limit value..is it 2r2 – user383053 Dec 20 '16 at 20:11
• What happens to $2r^2$ as $r$ goes to $0$? – David Dec 20 '16 at 20:11
• It's approaches 0 – user383053 Dec 20 '16 at 20:12
• That's basically the entire proof. For all $\theta$, as $r$ goes to zero we have that the function goes to $0$. Is something still troubling you about the problem? As a note: this only works because we don't require $\theta$ to be fixed, hence the function value approaches 0 independent of the path taken. – David Dec 20 '16 at 20:14