Removable and non-removable discontinuity in one function Is is possible to have a function with a removable and nonremovable discontinuity?
Is there a paper or site that I can see how this is possible or understand this better?
 A: Sure, you could have
$$ f(x) = \begin{cases} 0 & \text{when }x<0 \\
1 & \text{when }x=0 \\
0 & \text{when }0<x<1 \\
1 & \text{when }1\le x \end{cases} $$
which has a removable discontinuity at $0$ and a nonremovable one at $1$.
A: $$f(x) = \begin{cases} 
0, &  x = 0 \\ 
\frac{\sin(x)}{x}, & 0<x<1 \\
0, &x \geq 1 \\
\end{cases}.$$
The discontinuity at $0$ is removable since the limit of $\frac{\sin(x)}{x}$ as $x$ goes to $0$ is 1. The discontinuity at $1$ is not removable.
A: Consider 
$$f(x) = \begin{cases} 
1/x, &  x < 1 \\ 
0, & x = 1 \\
x, &x > 1 \\
\end{cases}.$$
Then $f$ has an non removable discontinuity at $x = 0$ (vertical asymptote), and a removable discontinuity at $x = 1$.
A: The function $$f(x)=\frac{x-1}{(x-1)(x-2)}$$ is not defined for $x=1$ and $x=2$.
However, the limit for $x\rightarrow 1$ exists.
$$\lim_{x\rightarrow 1} f(x)=\lim_{x\rightarrow 1} \frac{1}{x-2}=-1$$
Denominator and numerator tend to $0$. This does not imply that the
limit exists, but it is the case in this example. Setting
$f(1)=-1$, we can remove the singularity at $x=1$.
The limit for $x\rightarrow 2$ does not exist.
$$\lim_{x\rightarrow 2} f(x)=\pm \infty$$
depending on the way, x approaches to $2$.  The reason is, that the 
numerator remains positive and the denominator tends to $0$, if
$x$ tends to $2$. The singularity cannot be removed, no matter which
value we choose for $f(2)$.
