Let $f:\mathbb{R}^2 \to \mathbb{R}$ defined by f(0,0)=0 and f(x,y)=$\frac{x^2y}{x^4+y^2}$ if $(x,y)\ne(0,0)$
Is $f$ continuous at $(0,0)$?
$\vert\frac{x^2y}{x^4+y^2}\vert=\vert{y}\vert \vert\frac{x^2}{x^4+y^2}\vert \le\vert{y}\vert$ which tends to 0 as y$\to0$
Hence, by the sandwich theorem given function is continuous at (0,0).
Also along all the paths I am getting a limit of zero.
Please tell me any mistake.
Make the substitution $u=x^2$.
Now, you have $f(u,y)=\displaystyle \frac{uy}{u^2+y^2}$.
Converting to polar coordinates, $\displaystyle f(u,y)=f(r,\theta)=\frac{(r\cos(\theta))(r\sin(\theta))}{(r\sin(\theta))^2+(r\cos(\theta))^2}=\frac{r^2\sin(\theta)\cos(\theta)}{r^2}=\frac{1}{2}\sin(2\theta)$.
What does this even mean, you may ask?
Well, $\displaystyle \lim_{r \to 0}\frac{1}{2}\sin(2\theta)$ does not depend on $r$, and depends on $\theta$.
What does that mean? That means $f(r,\theta)=f(u,y)=f(x^2,y)$ approaches different values from different directions.
Therefore, the limit of $f(x,y)$ at $(0,0)$ does not exist.
Notice that the substitutions we made morph the function, but preserve the limit's properties.
