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?

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.

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 $\displaystyle \frac{x^2- 4}{x- 2}= \frac{(x- 2)(x+ 2)}{x- 2}= x+ 2$

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

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) $\displaystyle \lim_{x\to a} f(x)$ exists.

c) $\displaystyle \lim_{x\to a} f(x)= f(a)$.

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 $\displaystyle f(x)= \frac{x^2- 4}{x- 2}$ is not continuous at x= 2 but the function

"$\displaystyle g(x)= \frac{x^2- 4}{x- 2}$ 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.)