# Where the function is discontinuous.?

• Sep 1st 2011, 04:21 PM
rowdy3
Where the function is discontinuous.?
Find all values x=a where the function is discontinuous. For each value of x, give the limit of the function as x approaches a. Be sure to note when the limit doesn't exist.
f(x)= (x^2 -4) / (x-2)
The bottom I set it equal to zero and it's 2 DNE. In the back of the book it also list 4 as one of the answers. Where did the 4 come from? Is it from (x^2 - 4), I set it equal to zero and I get 4.
My teacher did an example f(x) (x^2 - 25) / (x+5) and she got x=-5. Wouldn't 25 be another answer?
• Sep 1st 2011, 04:35 PM
skeeter
Re: Where the function is discontinuous.?
Quote:

Originally Posted by rowdy3
Find all values x=a where the function is discontinuous. For each value of x, give the limit of the function as x approaches a. Be sure to note when the limit doesn't exist.
f(x)= (x^2 -4) / (x-2)
The bottom I set it equal to zero and it's 2 DNE. In the back of the book it also list 4 as one of the answers. Where did the 4 come from? Is it from (x^2 - 4), I set it equal to zero and I get 4.
My teacher did an example f(x) (x^2 - 25) / (x+5) and she got x=-5. Wouldn't 25 be another answer?

the question is asking the limit of the function where it is not continuous ... the "where" is at x = 2 ... the limit as x approaches 2 is 4.
• Sep 2nd 2011, 07:08 AM
HallsofIvy
Re: Where the function is discontinuous.?
Quote:

Originally Posted by rowdy3
Find all values x=a where the function is discontinuous. For each value of x, give the limit of the function as x approaches a. Be sure to note when the limit doesn't exist.
f(x)= (x^2 -4) / (x-2)
The bottom I set it equal to zero and it's 2 DNE. In the back of the book it also list 4 as one of the answers. Where did the 4 come from? Is it from (x^2 - 4), I set it equal to zero and I get 4.

I suspect you have misunderstood the question. "2" and "4" are answers to completely different questions.

First, you are correct that x= 2 is the only point at which the value of the function, as given, does not exist and so is the only value of x at which the function is not continuous. To find the limit, for x not equal to 2, we have $\frac{x^2- 4}{x- 2}= \frac{(x- 2)(x+ 2)}{x- 2}= x+ 2$

Because the limit of a function "as x goes to a" does NOT depend upon the value at a, $\lim_{x\to 2}\frac{x^2- 4}{x- 2}= \lim_{x\to 2} x+ 2= 4$.

That is, the "value x= a where the function is discontinuous" is x= 2. The "limit of the function", as x goes to 2, is 4.

Quote:

My teacher did an example f(x) (x^2 - 25) / (x+5) and she got x=-5. Wouldn't 25 be another answer?
You can't have an "answer" without a question! You say your teacher "did an example" but don't say what the question was that she was answering. It looks like your teacher was finding where f(x) is not defined or where it was not continuous. x= -5 is the only answer to that. IF the question were "what is the limit as x goes to -5", THEN the answer would be "25" but that is a different question.

By the way, you probably learned that a function, f, is continuous at x= a if and only if three things are true:
a) f(a) exists.
b) $\lim_{x\to a} f(x)$ exists.
c) $\lim_{x\to a} f(x)= f(a)$.

The two functions you give are discontinuous, the first at x= 2, the second at x= -5, because the function value does not exist there so (a) is not true. For these particular functions, (b) is true- the limit exists and is equal to 4 in the first case and 25 in the second. We could "make" the function continuous by defining the value at x= 2 and x= -5 to be that
limit.

That is, the function $f(x)= \frac{x^2- 4}{x- 2}$ is not continuous at x= 2 but the function
" $g(x)= \frac{x^2- 4}{x- 2}$ if [tex]x\ne 2[/itex], g(2)= 4" is continuous at x= 2. We say that f has a "removable discontinuity" at x= 2.

(You should be aware that this is a rather special situation. For most functions that are discontinous at "x= a", the limit itself does not exist.)