Does a limit exist at a point of discontinuity?

In summary: A function is continuous at a point if the limit as x approaches that point from both the left and right sides is equal to the actual value of the function at that point. In this case, since the limit as x approaches 1 from both sides is undefined, the function is not continuous at x= 1. Therefore, the statement is not true. In summary, a removable discontinuity is when the limit as x approaches a from both sides is equal but the actual value of the function at that point is not. This can be represented as a hole on the graph. In order for a function to be continuous at a point, the limit and the actual value must be equal. Therefore, the statement that a limit exists at a point
  • #1
grace77
43
0
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?
 
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  • #2
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
CAF123 said:
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.
ImageUploadedByPhysics Forums1392930823.848271.jpg


So let's say that it is asking does limf(x) at x-> exist for this graph, is the answer no?
 
  • #4
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
CAF123 said:
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?

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
The text that you quoted, grace77, said "[itex]\lim_{x\to a^-} f(x)= \lim_{x\to a^+} f(x)[/itex]" Do you not understand what that means? For x between 0 and 1, f(x)= 2x so [itex]\lim_{x\to 1-} f(x)= \lim_{x\to 1^-} 2x[/itex]. What is that equal to? For x between 1 and 2, f(x)= -2x+ 4 so [itex]\lim_{x\to 1^+} f(x)= \lim_{x\to 1^+} -2x+ 4[/itex]. What is that equal to? What does that tell you about [itex]\lim_{x\to 1} f(x)[/itex]?
 
  • #7
grace77 said:
Im just getting mixed up with it : is this statement true? A limit exists at a point of discontinuity but it is not continuous.
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) [itex]\lim_{x\to a} f(x)[/itex] exists.
3) [itex]\lim_{x\to a} f(x)= f(a)[/itex]
 

FAQ: Does a limit exist at a point of discontinuity?

What is a removable discontinuity?

A removable discontinuity is a type of discontinuity in a function where there is a hole or gap in the graph. This means that there is a point in the function where the limit exists, but the value of the function does not.

How is a removable discontinuity different from other types of discontinuities?

A removable discontinuity is different from other types of discontinuities, such as jump discontinuities or infinite discontinuities, because it can be removed by redefining the function at that point. This means that the limit of the function exists at the point of the discontinuity, but the function is not continuous at that point.

What causes a removable discontinuity?

A removable discontinuity is caused by a factor in the function that is undefined at a certain point. This could be due to a rational function having a common factor in the numerator and denominator that cancels out, or a piecewise function having a different definition at a specific point.

How can a removable discontinuity be removed?

A removable discontinuity can be removed by defining the function at the point of the discontinuity. This can be done by finding the limit of the function at that point and setting the value of the function to be equal to that limit. In other words, the function can be "filled in" at the point of the discontinuity to make it continuous.

Why is it important to identify and remove removable discontinuities?

Identifying and removing removable discontinuities is important because it allows for a more accurate representation of the function. By removing the discontinuity, the function becomes continuous and can be evaluated at that point. This is especially important in applications such as graphing and finding the derivative of a function.

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