# Continuity and differentiability of $f(x,y)$ at $(0,0)$

Define a function $f$ of two real variables in the $xy$ - plane by

$$f(x,y) =\begin{cases} \dfrac{x^3\cos\frac{1}{y} + y^3\cos\frac{1}{x} }{x^2+y^2}, &(x,y) \neq (0,0)\\ \\0, & (x,y) = (0,0)\end{cases}$$

Check the continuity and differetntaibility of $f$ at $(0,0)$

My approach: I tried first to check differentiability.

I first tried to calculate $f_x(0,0)$ using the formula $$\lim_{h\to0} \frac{f(h,0)-f(0,0)}{h}$$

I get $$\lim_{h\to0} \cos \frac{1}{0}$$ which oscillates bewteen $-1$ and $1$ and hence doesn't exist.

So $f_x(0,0)$ doesn't exist. Show $f$ is not differentiable at $(0,0)$.

For continuity,

I tried to use the formula $$\lim_{(x,y)\to(0,0)} f(x,y)$$ by approaching $(0,0)$ through $y=mx.$ But I couldn't solve this limit.

Please tell me if my argument on differentiability is correct or not. and how to check continuity.

• how is $f(x,0)$ defined? – Guy Fsone Oct 21 '17 at 14:18
• The simple answer is that $f$ is not defined in any full neighborhood of $(0,0),$ hence $f$ cannot be differentiable there. – zhw. Oct 21 '17 at 17:45

Your Argument for the differentiability seems ok but I wounder how do you define $f(x,0)$
$$|x|\le \|(x,y)\|~~~and~~~|y|\le \|(x,y)\|~$$ since $\|(x,y)\|^2 = x^2+ y^2$.
Hence, $$\left|\frac{x^3cos\frac{1}{y} + y^3cos\frac{1}{x} }{x^2+y^2}\right| \le \frac{|x|^3+ |y|^3}{\|(x,y)\|^2} \le \frac{2\|(x,y)\|^3}{\|(x,y)\|^2}\le 2\|(x,y)\|\to 0$$