# Removable discontinuity

1. Feb 20, 2014

### grace77

Just wanted to clear up something :
If there is a removable discontinuity on a graph(1,2) with both a right and left hand portion and let's say the question was : whether lim at x-> 1f(x) exist

Would it be yes it does exist just that it is undefined or?

2. Feb 20, 2014

### CAF123

A removable discontinuity is one in which $\lim_{x \rightarrow a^-} f(x) = \lim_{x \rightarrow a^+} f(x) = L$, but $f(a) \neq L$. So on a graph, you represent the discontinuity as a hole, and the actual value f(a) as a dot. It means that the value $a$ is not in the domain of the original function $f(x)$, but by declaring that $f(a) = L$, the function can be made continuous.

3. Feb 20, 2014

### grace77

So let's say that it is asking does limf(x) at x-> exist for this graph, is the answer no?

4. Feb 20, 2014

### CAF123

Why do you think no? In more simplistic terms, as you approach 1 from the left, what value is f(x) tending to? As you approach 1 from the right, what value of f(x) are you tending to?

5. Feb 20, 2014

### grace77

Im just getting mixed up with it : is this statement true? A limit exists at a point of discontinuity but it is not continuous.

6. Feb 20, 2014

### HallsofIvy

The text that you quoted, grace77, said "$\lim_{x\to a^-} f(x)= \lim_{x\to a^+} f(x)$" Do you not understand what that means? For x between 0 and 1, f(x)= 2x so $\lim_{x\to 1-} f(x)= \lim_{x\to 1^-} 2x$. What is that equal to? For x between 1 and 2, f(x)= -2x+ 4 so $\lim_{x\to 1^+} f(x)= \lim_{x\to 1^+} -2x+ 4$. What is that equal to? What does that tell you about $\lim_{x\to 1} f(x)$?

7. Feb 20, 2014

### HallsofIvy

The limit exists at the point of discontinuity but the function is not continuous there.

In order that f(x) be continuous at x= a, three things must be true:
1) f(a) exists.
2) $\lim_{x\to a} f(x)$ exists.
3) $\lim_{x\to a} f(x)= f(a)$