Define $f: \Bbb R^2 \to \Bbb R$ by $f(x,y) =\begin{cases} \frac{xy^2}{x^2 + y^4}, & \text{if $(x,y) \ne 0$} \\ 0, & \text{if $(x,y)= 0$} \end{cases}$ Define $f: \Bbb R^2 \to \Bbb R$ by
$f(x,y) =\begin{cases}
\frac{xy^2}{x^2 + y^4},  & \text{if $(x,y) \ne 0$} \\
0, & \text{if $(x,y)= 0$}
\end{cases}$
I'm to show that $f$ is continuous at any straight line THROUGH the origin, but not continuous AT the origin. I'm confused how this is possible. Is it a hole?
How would I do this?
 A: To show that it's continuous along straight lines:
Approach the origin along the half line $(x,y) = (r\cos\theta,r\sin\theta)$, where $r>0$ and $r\to 0^+$.
\begin{eqnarray*}
\frac{xy^2}{x^2+y^2} &\rightsquigarrow& 
\frac{r^3\cos\theta\sin^2\theta}{r^2\cos^2\theta+r^2\sin^2\theta} \\ \\
&\equiv& \frac{r^3\cos\theta\sin^2\theta}{r^2} \\ \\
&\equiv& r\cos\theta\sin^2\theta
\end{eqnarray*}
As $r\to 0$, $r\cos\theta\sin^2\theta \to 0$ for all $\theta$.
A: Let $$f(x,y) =\begin{cases}
\frac{xy^2}{x^2 + y^4},  & \text{if $(x,y) \ne 0$} \\
0, & \text{if $(x,y)= 0$}
\end{cases}$$
Now observe that for $(c_1,c_2)\neq (0,0)$ the map $t\mapsto (t c_1, t c_2)$ define an arbitrary straight line that pass by the points $(0,0)$ and $(c_1,c_2)$ and we find that
$$\frac{xy^2}{x^2+y^4}=\frac{tc_1c_2^2}{c_1^2+t^2 c_2^4}=\frac{c_1 c_2^2}{\frac{c_1^2}t+t c_2^4}\to 0,\quad\text{as } t\to 0$$
Then $f$ is continuous through straight lines that pass by the origin. Now to show that $f$ is not continuous at $(0,0)$ is enough to make the change $x=y^2$, then
$$\lim_{(y^2,y)\to(0,0)}f(y^2,y)=\lim_{y\to 0}\frac{y^4}{2y^4}=\frac12\neq 0$$
A: $$f(x,y) =\begin{cases}
\frac{xy^2}{x^2 + y^4},  & \text{if $(x,y) \ne 0$} \\
0, & \text{if $(x,y)= 0$}   \end{cases} $$
Note that ax $x\to 0$, 
along straight lines 
$$ y=mx \implies f(x,mx)= \frac{m^2x}{1+m^4x^2} \to 0$$
along another curve, 
$$ y=\sqrt x \implies f(x,\sqrt x)= \frac{x^2}{x^2+x^2} \to 1/2$$
Thus the limit at $(0,0)$ does not exist. 
A: Note
$$   \left |\frac{xy^2 }{x^2+y^2}\right| = |y|\left |\frac{xy }{x^2+y^2}\right| \le |y|/2.$$
We have used the basic inequality $|xy| \le (x^2+y^2)/2$ here. It follows that the limit at $(0,0)$ is $0,$ so $f$ is continuous at $(0,0).$ Thus the problem is in error.
However, it is possible to find a function $g$ continuous along every straight line through $(0,0),$ with $g(0,0)=0,$ such that $g$ is not continuous at $(0,0).$ 
