removable discontinuity

I am having some confusions about the removable discontinuities. most of the functions with removable discontinuity I have faced are piece wise functions.example: $f(x)=\begin{cases}x+2,x\neq 1 \\4,x=1\end{cases}$

my first question:how can the discontinuity of a piecewise function be removed??

2nd question:functions with removable discontinuities usually violate the 3rd condition of continuity,which is understandable for piecewise functions.but the function $f(x)=\frac{(x-1)(x-2)}{(x-1)}$ also has a removable discontinuity at $x=1$.but this function violates the very first condition of continuity ($f(a)$ is not defined).should it still be called a removable discontinuity??

• sorry I'll edit the function.it was not meant to be continuous Jun 5, 2016 at 20:14
• and my 1st question was how can the discontinuity of any piecewise function can be removed Jun 5, 2016 at 20:15

It doesn't matter which of the three conditions of continuity are violated, as long as the second condition is met while the third is not, we have a removable discontinuity.

If the second is not met, then regardless of the first we have a nonremovable discontinuity.

I'm assuming your conditions are these:

1. $f(c)$ is defined
2. $\lim_{x\to c}$ exists
3. Limit and actual value are identical at $x=c$
• Because $\lim_{x\to1}f(x)\neq f(1)$ (I.e. because conditions 1 & 2 are met but 3 is not) Jun 5, 2016 at 20:32
A function has a removable discontinuity at $x_0=a$, if and only if the limit for $x\rightarrow a$ exists (left and right limit must coincide!) and if it is not continous at $x_0=a$ (either not defined or with the wrong value).
To remove a discontinuity, either add or change a point (in the first example , change $(1/4)$ to $(1/3)$)