I am looking for a verification of my solution to this problem. I have to determine whether $$f(x,y)=\begin{cases}\frac{x^2\sin{y^2}}{x^2+y^4} \quad &(x,y)\in\mathbb{R}^2\setminus\{(0,0)\}\\0 &(x,y)=(0,0)\end{cases}$$ is continuous and differentiable in the origin.
So first observe $|f(x,y)|=|\frac{x^2\sin{y^2}}{x^2+y^4}|\leq|\frac{\sin{y^2}}{1+y^4/x^2}|\leq|\sin{y^2}|\to0$ as $(x,y)\to0$, so that $f$ is continuous at the origin.
Next, it is easy to show that $f_x(0,0)=\lim_{t\to0}\frac{f(t,0)-f(0,0)}{t}=\lim_{t\to0}\frac0t=0$ and likewise $f_y(0,0)=0$. So clearly the candidate for the total derivative would be $\begin{bmatrix}0&0\end{bmatrix}$.
So the limit I want to calculate is $\lim_{(x,y)\to(0,0)}\frac{f(x,y)}{||(x,y)||}$ and if we calculate for $x\neq0$ $$\left|\frac{f(x,y)}{||(x,y)||}\right|=\left|\frac{x^2\sin(y^2)}{(x^2+y^4)\sqrt{x^2+y^2}}\right|\leq\left|\frac{x^2y^2}{(x^2+y^4)\sqrt{x^2+y^2}}\right|\leq\left|\frac{x^2y^2}{x^2\sqrt{x^2+y^2}}\right|,$$ we observe that this last fraction is homogeneous to degree $1$, thus this fraction and therefore $\frac{f(x,y)}{||(x,y)||}$ go to $0$ as $(x,y)\to(0,0)$. Thus $f$ is indeed differentiable in the origin and the total derivative is given by $\begin{bmatrix}0&0\end{bmatrix}$.
My question: am I right in these claims? When I enter $\lim_{(x,y)\to(0,0)}\frac{x^2\sin(y^2)}{(x^2+y^4)\sqrt{x^2+y^2}}$ into Mathematica, the output is the same as the input. This made me doubt and I'm not sure about the total differentibality, also since other results did not apply here, e.g. the partial derivatives do not seem continuous in the origin (I know this is only a sufficient and not a necessary condition). Also, does anyone know why Mathematica does not yield the right output (either the limit value or 'indeterminate') in this case? For example when I enter $y$ without a square in the limit in Mathematica I do get 'indeterminate'.
Everything is fine. I wouldn't have used homogeneity, but that argument works as long as the function is bounded on the unit circle, which is true here (although it should be explained). I would have used
$$\frac{x^2y^2}{(x^2+y^4)(x^2+y^2)^{1/2}}= \frac{x^2}{(x^2+y^4)}\frac{y^2}{(x^2+y^2)^{1/2}} \le 1\cdot \frac{x^2 +y^2}{(x^2+y^2)^{1/2}} = (x^2+y^2)^{1/2} \to 0.$$