0
$\begingroup$

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?

$\endgroup$
  • 2
    $\begingroup$ Pick the polar coordinates $x=r\cos\theta$, $y=r\sin\theta$ and see what happens for $(x,y)\to (0,0)\iff r\to 0$. $\endgroup$ – A.Γ. Jun 22 at 16:29
  • $\begingroup$ We have not used them yet in my class so I unfortunately have very little idea what you mean. Is there a way to prove it without using them? $\endgroup$ – math_beginner Jun 22 at 16:32
  • 2
    $\begingroup$ $0\leq(x-y)^2\implies xy\leq {x^2+y^2\over2}$ $\endgroup$ – saulspatz Jun 22 at 16:45
1
$\begingroup$

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 $(0,0)$.

$\endgroup$
0
$\begingroup$

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, $$\lim_{(x,y)\to(0,0)}f(x,y)=0.$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.