Is the function $\frac { x-1 }{ \ln { { \left( x \right) }^{ 2 } } } $ continuous at$ x=0$? I would like to know whether the function shown in the title is continuous or not at $x=0$.
This problem is disturbing since the function isn't defined at x=0, but the limit of the function as x approaches to $0$ is $0$.
Thanks!
 A: There is no definition of continuity outside the domain of a function. I mean no offense, but to explain: asking if a function is continuous at a point that is not in its domain is like asking if the function is green.
A: As written, the function isn't well-defined at $0$ so the question doesn't really make sense. What you're asking is similar to asking if $f: [0,\infty) \to [0,\infty)$  defined by $f(x) = \sqrt x, x \ge 0$ is continuous at $x=-1$. 
By contrast, you could define $f$ by $$f(x) = \left\{\begin{matrix} \frac{x-1}{\log(x^2)}, & x\neq 0,1, \\0, &x=0,1. \end{matrix} \right.$$ (The function also isn't well-defined at $x=1$ which is why I had to add that as well.) This function is well-defined everywhere and is continuous at $x=0$.
A: The function $\frac{x-1}{\log(x^2)}=\frac{x-1}{2\log(x)}$ has a removable discontinuity at $x=1$.  That is, we can define a function $f(x)$ such that
$$f(x)=\begin{cases}\frac{x-1}{\log(x^2)}&,x\ne 1\\\\0&,x=1 \tag 1\end{cases}$$
that is continuous at $x=1$.

However, the $\lim_{x\to 0}\frac{x-1}{\log(x^2)}=\infty$, and therefore, $f(x)$ is not continuous at $x=0$ and does not have a removable discontinuity at $x=0$.

