Is the function $f(x, y)= \frac{x^3}{x³+y^2}$ continuous? I want to know if the function is continuous.
$$f(x,y)=\begin{cases}\frac{x^3}{x^3+y^2},&(x,y)\neq(0,0)\\0,&(x,y)=(0,0)\end{cases}$$
First of all, I evaluate the limit by putting the point $(0,0)$ into the function:
$$\lim_{(x,y)\to (0,0)}\frac{x^3}{x^3+y^2}$$
?
The result is 0/0.
Then, I used the path rule.
The first path: $y=x$
The result is $0$.
The second path is $y=x^2$. The result is $1$.
The third path is $y=0$. The result is $1$.
Are my result correct? The book says that this limit exists.
 A: Your proof with the paths $y=x^2$ and $y=x$ is correct and shows that the limit at the point $(0,0)$ does not exist.
There is also another way to prove this and it is quite useful for someone to keep in mind.
Take $$x=r\cos{\theta}$$ $$y=r\sin{\theta}$$
We have that $\lim_{(x,y) \longrightarrow (0,0)}\frac{x^3}{x^3+y^2}=\lim_{r \longrightarrow 0} \frac{r^3cos^3 \theta}{r^3cos^3 \theta+r^2 sin^2 \theta}=\lim_{r \longrightarrow0}\frac{rcos^3 \theta}{rcos^3 \theta+sin^2 \theta}$
This last  limit, in order to exist,  must be independent of $\theta$.
In a first glance someone might say that when $r$ approaches $0$ the limit is $0$,but if you notice this limit is not independent of the values of $\theta, \forall \theta \in [0,2\pi]$(we can let $\theta$ take values in all the real line but it suffices to use the interval $[0,2\pi]$ because $cos \theta$ and $sin \theta$ are $2 \pi-$periodic functions)   because different values of $\theta$ give different results
If you take $\theta= \pi/2$ then $\lim_{r \longrightarrow0}\frac{rcos^3 \theta}{rcos^3 \theta+sin^2 \theta}=0$, but for $\theta=\pi$ we have that
$\lim_{r \longrightarrow0}\frac{rcos^3 \theta}{rcos^3 \theta+sin^2 \theta}=1$
Thus the limit does not exist.
In general to compute a limit near a point $(a,b)$ you can put $x=a+rcos \theta$ and $y=b+rsin \theta$ and take the limit when $r$ approaches $0$.
Then we have our limit when the result is independent of $\theta$
