The title is rather vague, the question is more along the line of, does there exist a number $c$, such that
$$f(x,y) = \begin{cases}
\frac{2xy+x^4}{x^2+y^2} & (x,y)\neq (0,0)\\
c & (x,y)=(0,0) \end{cases} $$
is continuous.
So I was thinking if I take the limit and see what the upper part of the piecewise function approaches, that $c$ would have to be something along does lines, like you do in single-variable calculus. So I went about my day and parametrized $x=r\cos\theta $ and $y=r\sin\theta$
Such that I'd get
$$\lim_{(x,y)\rightarrow (0,0)} \frac{2r^2\sin\theta\cos\theta+r^4\cos^4\theta }{r^2(\sin^2\theta+\cos^2\theta)}=\lim_{(x,y)\rightarrow (0,0)}\frac{r^2(\sin2\theta +r^2\cos^4\theta)}{r^2}=\sin2\theta $$ Which obviously is path dependent, so the limit doesn't exist, but whenever I smash values obtusely close to $(0,0)$ into my calculator it approximates the functional value to $1$. So how can I solve this?
Best Answer
It's usually easier to find limits along lines by inserting either $y = kx$ or $x = ky$ for some constant $k\in \Bbb R$, than transforming to polar coordinates. That way we get, for $x\neq 0$, $$ f(x, kx) = \frac{2x\cdot kx + x^4}{x^2 + (kx)^2} = \frac{2kx^2 + x^4}{x^2 + k^2x^2} = \frac{2k + x^2}{1+k^2} $$ whose limit as $x\to 0$ clearly depends on $k$ (as a bonus, this coupled with your answer gives a relationship between the slope $k$ of a line and its angle $\theta$: $\frac{2k}{1+k^2} = \sin2\theta$). In particular, for $k = 1$ the limit is $1$, and for $k = -1$ the limit is $-1$. So no value for $c$ will make the function continuous.
Beware, however. Some functions have the same limit at the origin along any line, yet don't have a limit at the origin. For instance $$ f(x, y) = \cases{\frac{x^2y}{x^4 + y^2} & for $(x, y)\neq(0,0)$\\ 0 & for $(x, y) = (0,0)$} $$ has limit $0$ at the origin along any line, but along the parabola $y = x^2$ we have the limit $\frac12$.