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?
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$\begingroup$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 $1$.
$\endgroup$ $\begingroup$$$f(x) = \begin{cases} 0, & x = 0 \\ \frac{\sin(x)}{x}, & 0<x<1 \\ 0, &x \geq 1 \\ \end{cases}.$$
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.
$\endgroup$ $\begingroup$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 $f(2)$.
$\endgroup$ 1 $\begingroup$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 $x = 1$.
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