Show that multivariable function is continuous but not differentiable

Question

Show that $f$ is continuous and in neigbourhood of $(0,0)$ has bounded partial derivatives but is not differentiable. $$ f(x,y) = \begin{cases} \frac{xy}{\sqrt{x^2+y^2}} & x^2+y^2 \neq 0 \\ 0 & x = y = 0 \end{cases} $$ This seems like basic introductory material and is also a copy of the same question on StackExchange, but unfortunately it was not sufficiently answered there.

I'm stuck at the beggining. Function f is said to be continous at point (0,0) when it satisfies the following $$ \lim_{(x_0,y_0) \rightarrow (0,0)}f(x,y)= f(0,0) $$

When I plug that I immediately get: $$ \lim_{(x_0,y_0) \rightarrow (0,0)}f(x,y) = \lim_{(x_0,y_0) \rightarrow (0,0)} \frac{x_0y_0}{\sqrt{x_0^2 + y_0^2}} = \frac{0}{0} = \text{ ?} $$ How should I tackle that problem?

Answer 1

Let $x=r\cos \theta, \quad y=r\sin \theta$

Therefore $$|\frac{xy}{\sqrt{x^2+y^2}}|=r |\cos \theta \sin \theta|\le r=\sqrt{x^2+y^2}\lt \epsilon$$if $x^2\lt\frac{\epsilon^2}{2},\quad$and$\quad y^2\lt \frac{\epsilon^2}{2}$

or if $|x|\lt\frac{\epsilon}{\sqrt2},\quad$and$\quad |y|\lt \frac{\epsilon}{\sqrt2}$

Thus $$|\frac{xy}{\sqrt{x^2+y^2}}-0|\lt \epsilon,\qquad \text{where}\quad |x|\lt\frac{\epsilon}{\sqrt2},\quad and \quad |y|\lt \frac{\epsilon}{\sqrt2}$$

$$\implies \lim_{(x,y)\to (0,0)}\frac{xy}{\sqrt{x^2+y^2}}=0$$

Hence $$\lim_{(x,y)\to (0,0)}f(x,y)=f(0,0)$$and therefore $f(x,y)$ is continuous at $(0,0)$

Again $$f_x(0,0)=\lim_{h \to 0}\frac{f(h,0)-f(0,0)}{h}=0$$and $$f_y(0,0)=\lim_{k \to 0}\frac{f(0,k)-f(0,0)}{k}=0$$

Thus the function $f(x,y)$ possesses partial derivatives at $(0,0)$.

If the given function is differentiable at $(0,0)$, then by definition $$df=f(h,k)-f(0,0)=Ah+Bk+h\phi +k\psi\qquad . . . . . (1)$$ where $\quad A=f_x(0,0)=0;\quad B=f_y(0,0)=0, \quad $and $~\phi,~\psi~$ tends to zero as $\quad (h,k)\to (0,0)$.

So from $(1)$ we have $$\frac{hk}{\sqrt{h^2+k^2}}=h\phi +k\psi$$

Putting $\quad k=mh \quad $and letting $\quad h\to 0,\quad$ we have

$$\frac{m}{\sqrt{1+m^2}}=\lim_{h \to 0}(\phi +m\psi)=0$$which is impossible for arbitraty $~m$.

Hence the function is not differentiable at $(0,0)$.

Answer 2

First, remember that $$\textrm{If } \lim_{(x,y)\to(0,0)}|f(x,y)|=0,\ \mbox{then} \lim_{(x,y)\to(0,0)}f(x,y)=0.$$

Now, by the inequality of arithmetic and geometric means, we have $$|xy|=|x||y|\leqslant\frac{x^2+y^2}{2}=\frac{(\sqrt{x^2+y^2})^2}{2}\Rightarrow\frac{|xy|}{\sqrt{ x^2+y^2}}\leqslant\frac{\sqrt{x^2+y^2}}{2}.$$ Thus, since $|f(x,y)|\geqslant0$, we have $$0\leqslant\lim_{(x,y)\to(0,0)}|f(x,y)|=\lim_{(x,y)\to(0,0)}\left|\frac{xy}{\sqrt{x^2+y^2}}\right|\leqslant\lim_{(x,y)\to(0,0)}\frac{\sqrt{ x^2+y^2}}{2}=0.$$ Therefore, $$\lim_{(x,y)\to(0,0)}|f(x,y)|=0.$$ Hence, $$\lim_{(x,y)\to(0,0)}f(x,y)=0.$$