is the limit continuous or not? is $(x^2 + y^2) / (x^2 - y^2)$ continuous or not at $(0,0)$?
I think it is not continuous at $(0,0)$. To check I just plugged in the points and got $0$. Did I do that right? and also is there another way to do this?
 A: In order that a function be continuous at a point, it must be defined at that point. So, if the given function is
$$
f(x)=\frac{x^2+y^2}{x^2-y^2}
$$
without any other specification, continuity at $(0,0)$ is out of the question because there's no defined value for the function at $(0,0)$.
The function is, however, continuous everywhere it's defined, since it's a quotient of continuous functions.
The problem is then whether $f$ can be extended to a continuous function on $\mathbb{R}^2$ which, in this case, is equivalent to ask whether
$$
\lim_{(x,y)\to(a,a)}\frac{x^2+y^2}{x^2-y^2}
$$
exists (and is finite), for any $a$, because the function is defined on all points $(x,y)$ with $x\ne y$. For $a=0$ the limit doesn't exist: indeed
$$
\lim_{y\to0}\frac{0^2+y^2}{0^2-y^2}=-1
$$
while
$$
\lim_{x\to0}\frac{x^2+0^2}{x^2-0^2}=1
$$
For $a\ne0$, one can do in this way:
$$
\lim_{h\to0}f(a+h,a)=\lim_{h\to0}\frac{2a^2+2ah+h^2}{2ah+h^2}=\pm\infty
$$
where $\infty$ or $-\infty$ must be chosen depending if $a>0$ or $a<0$. Thus $f$ cannot be extended to a continuous function also at points $(a,a)$ for $a\ne0$.
A: If we approach $(0,0)$ along the $x$-axis (so $y=0$), we find the limit is $1$. If we approach along the $y$-axis so that $x=0$, we get the limit to be $-1$. What does this tell us about continuity?
A: More extending approach may be to consider a continuous curve: $$C_{\alpha}:t\longrightarrow\alpha t$$ in $\mathbb R^2$. Using it we have $$\lim_{t\to 0}f(C_{\alpha}(t))=\frac{\alpha^2+1}{-\alpha^2+1}$$ which is clearly dependent to value of $\alpha$. This shows what other answers confirmed.
