# Removable discontinuity

1. Nov 9, 2005

### maria curie

the book says that g(x)= x ,if x is not equal to 2 / 1,if x=2
has a removable disc. at x =2.

I couldn't remove it.I guess I didn't understand a removable dic. completely.I have an exam on friday.I need your help

thanks

2. Nov 9, 2005

### HallsofIvy

Staff Emeritus
f(x)= x if x is not equal to 2
f(x)= 1 if x is equal to 2.

What is the limit of f, as x-> 2? (Remember that the limit depends only on the values of f close to 2, not at 2!)

WHY is f not continuous at x= 2??

What would happen if you changed the value of f(2) to 2?

3. Nov 9, 2005

### maria curie

f(x) is not cont at x=2 because f(2)=1 is not equal to lim x->2f(x)=2.I know what continuity is.I dont know how we change the value of f(2) to 2 to be continuous x=2??

and what are the conditions of the continuous extension ?Can be every funct.that is not continuous become cont.?

Last edited: Nov 9, 2005
4. Nov 9, 2005

### 1800bigk

a removable discontinuity is just something that we can fix or adjust to get the function continuous. It usually means a function is discontinuous at some point or hole in the graph and all we have to do is plug the hole if you will, or redefine the function at the point in question. The problem you have has removable discontinuity because all we have to do is redefine the function at some point to get continuity.

1/x has an infinite discontinuity at zero and cannot be fixed

5. Nov 9, 2005

### HallsofIvy

Staff Emeritus
A function is continuous at x= a if and only if:
1) f(a) exists
2) $lim_{x\rightarrowa}f(x)$ exits.
3) $lim_{x\rightarrowa}f(x)= f(a)$

The function as given satisfies the first two of those at x= 2. It is not continuous because the third is not true: $lim_{x\rightarrow2}f(x)= 2$ but f(2)= 1. Redefining f(2)= 2 will make all three true! That "removes the discontinuity". (In fact, the function becomes f(x)= x for all x.)

As long as the limit exists at x= a we could always redefine f(a) to be that limit and "remove the discontinuity". A discontinuity is "not removeable" if the limit does not exist. $f(x)= \frac{1}{x}$ is not continuous at x= 0, as 1800bigk says, and no value for f(0) will make it continuous: it has an "infinite discontinuity" there.
The function: f(x)= x if x<= 1, f(x)= x+ 2 if x> 1, has a "jump" discontinuity at x= 1. The "limit from the left" is 1, the "limit from the right" is 3. Since those are not the same, there is no "limit" at x= 1.

6. Nov 9, 2005

### CarlB

Marie, this problem is a lot simpler than you're making it. What you're looking for is a function f(x) that is similar to g(x), but is more continuous. The rules are that you must have f(x) be the same as g(x) when x is not 2, but you can choose whatever you like for f(2). Now, take my word, the simplest thing for f(x) that comes to your mind is the right answer.

A lot of the time, mathematics is about finding a really complicated way of saying something that is very simple. Don't walk around thinking that this stuff is too complicated, it's not. It's just that it's not being explained to you in simple terms.

Carl

7. Nov 10, 2005

### maria curie

HallsofIvy ,your explanation is very good,thanks a lot