Question

Redefine the function f(x) =[x]+[x] in such a way that it becomes continuous for $x\in (0,2)$

Answer 1

Here ${lim}_{x\to 1}f\left(x\right)=-1$ but f(a)=0
Hence, f(x) has a removable discontinuity at x =1.
To remove this we define f(x) as follows
$f\left(x\right)=\left[x\right]+[-x],x\in (0,1)\cup (1,2)=-1,x=1.$
Now,f(x)is continuous for $x\in (0,2)$
Missing Point Discontinuity
The discontinuity is said to be of missing point type if the limit of the function exists at point 'a'

Consider the function $f\left(x\right)=\frac{x_2-4}{x-2}$, where $x\ne 2$
It is visible from the graph as well as the definition of the function
Isolated Point DIscontinuity
In this type of discontinuity, not only does the limit of the function
Example; Consider the function
f(x)=sgn (cos2x -2sin x +3) =sgn (2(2+sin x)(1-sinx))=0, if $x=2n\pi +\frac{\pi }{2}$
$=+1,ifx\ne 2n\pi +\frac{\pi }{2}$ has an isolated point at x =0 discontinuity as $x=2n\pi +\frac{\pi }{2}$