da javicemarpe » 30/03/2017, 14:32
The continuity doesn't follow from $|\sin(x)|\leq |x|$ but from $|x\sin(1/x)|\leq |x|$.
On the other hand, if you want to study the differentiability of the function $f$ on $(0,0)$ you have to find one (the only one) linear application $L$ such that $$\lim_{(x,y)\to(0,0)}\frac{f(x,y)-f(0,0)-L[(x,y)-(0,0)]}{\|(x,y)\|}=0.$$
I'll show you the step-by-step way to do it:
This $L$ is completely determined by the partial derivatives $\partial_x f(0,0)$ and $\partial_y f(0,0)$ in the following way: $L(x,y)=\partial_xf(0,0)x+\partial_yf(0,0)y$. So the only thing you have to do is to compute these derivatives.
Now, $$\partial_x f(0,0)=\lim_{h\to0}\frac{f(h,0)-f(0,0)}{h}=\lim_{h\to 0}\frac{0}{h}=0,$$
and the same happens with $\partial_yf(0,0)$. Then, the only candidate to be the differential of $f$ at $(0,0)$ is the null function $L(x,y)=0$.
Let's check it. First, let us observe that the candidate to be the limit is $0$ because, when $x=0$ or $y=0$, the function is identically $0$ and so is the limit of the definition of differentiability. So, we have to check if it is true that
$$0=\lim_{(x,y)\to(0,0)}\frac{f(x,y)-f(0,0)-L(x,y)}{\|(x,y)\|}=\lim_{(x,y)\to(0,0)}\frac{f(x,y)}{\|(x,y)\|}$$
even when we approach the point $(0,0)$ with points $(x,y)$ such that $xy\ne 0$.
But, if $xy\ne 0$,
$$\left|\frac{f(x,y)}{\|(x,y)\|}\right|=\left|\frac{xy\sin\left(\frac{1}{xy}\right)}{\|(x,y)\|}\right|\leq \frac{|xy|}{\|(x,y)\|}\leq \frac{|x|\|(x,y)\|}{\|(x,y)\|}=|x|,$$ which goes to $0$ when $(x,y)\to 0$. When $xy=0$, the function is identically $0$, so for these points is also clear that the limit is $0$. Then, the function is differentiable at $(0,0)$ and its differential is $L(x,y)=0$.
This is the step-by-step way to do it. If you check your final argument, you'll realise that what you have is that
$$\left|\frac{xy}{\sqrt{x^2+y^2}}\right|\leq\left|\frac{|\rho^2\cos\theta\sin\theta|}{\rho}\right|\leq \rho\to0.$$
Probably you wanted to say that thing. Anyways, you have to be careful with this kind of inequalities.