Prove that if $f(x, y)$ is continuous on $(0,0)$ then the function $g(x,y)=xyf (x,y)$ is differentiable on $(0,0)$

Question

I made the following: $$\lim_{(x,y)\rightarrow(0,0)} \frac{xy f(x,y) - 0 -g_{x}(0,0) - g_{y}(0,0) (y-0)}{((x-0)^2 + (y-0)^2)^{\frac{1}{2}}}$$ $$=\lim_{(x,y)\rightarrow(0,0)} \frac{xy f(x,y)}{(x^2+y^2)^{\frac{1}{2}}}$$

Now, my problem is how to continue to finish the demonstration.

Answer 1

You need to find $\lim_{(x,y)\to(0,0)}\frac{xyf(x,y)}{\sqrt{x^2+y^2}}$. Now the easiest way to find the limit in my opinion is change to polar coordinates. So $x=r\cos\theta,y=r\sin\theta$ and now the limit is:

$\lim_{r\to 0}\frac{r^2cos(\theta)sin(\theta)f(r\cos(\theta),r\sin(\theta))}{r}=\lim_{r\to 0}rf(r\cos(\theta),r\sin(\theta))cos(\theta)sin(\theta)=0$

The limit is zero because $rf(r\cos(\theta),r\sin(\theta))$ is a function which goes to zero when $r\to 0$ and $\cos(\theta)\sin(\theta)$ is a constant function with respect to $r$, so their product goes to zero without dependence on $\theta$.

Answer 2

Your last limit is just

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

because the right factor exists by hypothesis, and the left factor is $0$.