Let $$(f :\mathbb{R}^2 \to \mathbb{R})$$ be defined by $$\frac{x_1x_2^2}{(x_1^2+x_2^2)}$$ if $x$ $\neq(0,0)$, $f(0)$ = $0$.
Show that $f$ is continuous at $(0, 0)$.
Show that the partial derivatives $\frac{∂f}{∂x1}(0, 0),$ $\frac{∂f}{∂x2} (0, 0)$ exist and find their values.
(c) By using the definition ∗ show that f is not differentiable at $x$ = $0$.
Note that $0$ and $x$ are vectors
The * from above : $\lim_{x\to a} \frac{||F(x) - F(a)-DF(a)(x-a)||}{||x-a||} = 0.$
For a) I know the limit is 0, but wouldn't that mean the function is not continuous at 0? Or is it just 0? For b) if I take the partial's won't I just end up with 0? Not sure how to do c.
Any help is appreciated!
For part (c) since the partial derivatives are all zero, the formula reduces to proving that $$\lim_{(x,y)\rightarrow(0,0)}\frac{|f(x,y)|}{\sqrt{x^2+y^2}}\ne 0$$ Going to polar cooridinates we can write the expression as $$ \frac{r\cos(\theta)r^2\sin^2(\theta)}{r^3}=\cos(\theta)\sin^2(\theta)$$ So we see that the limit is not zero in any direction except along the $x$ or $y$ axis, and thus the limit does not exist.