# Thread: Removable Discontinuity Experts Required

1. ## Removable Discontinuity Experts Required

I'm not sure if I fully understand the removable discontinuity concept. Can someone check this over for me please?

f(x)=

{2cos(x/2), x<0
{ln(x), for 0<x<1
{2x^3 - x - 1, for x>1

I know that the lim of 2cos(x/2) as x-->0- = -2
I know that the lim of 2x^3 - x - 1 as x-->0+ = -16pi^3 +2pi - 1
So is it correct to say that the lim of ln(x) as x approaches 0 from the right DNE because ln(x) is undefined when x = 0

Therefore f(x) is discontinuous at x = 0, because the function is undefined at x = 0 and because the limit does not exist.

Would it also be correct to say that this function is an example of a removable discontinuity, if we redefine f(x)
to be

{2cos(x/2), x<0
{-2, 0<x<1
{-2, x>1

How many changes are you allowed to make to a function f(x) in order to make it into a removable discontinuity?

2. 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