Several things are incorrect in your thinking on this one

To have a removable discontinuity at a point the limit must exist at that point

For example if f(x) = sin(x)/x then f(0) is undefined

However lim sin(x)/x =1

x->0

Therefore if we define f(x) = {sin(x)/x if x is not 0

{1 if x = 0

then f(x) is continuous at 0

1. lim2cos(x/2) = 2 not -2

x->0-

2. lim ln(x) = - infinity

x->0+

Therefore limf(x) DNE

x->0

Therefore the discontiuity is non-removable

at x= 1

lim ln(x) = 0 and lim 2x^3 - x - 1 = 0

x->1 x->1

therefore define f(x) = 0 at x= 1 to remove the discontiuity.

Also if we define

{ ln(x), for 0<x<1

{2x^3 - x - 1, for x>1

or

{ ln(x), for 0<x<1

{2x^3 - x - 1, for x>1

then we will have removed the discontinuity since both ln(x) and

2x^3 - x -1 are defined at x = 1