If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

## AP®︎/College Calculus AB

### Course: AP®︎/College Calculus AB>Unit 1

Lesson 13: Removing discontinuities

# Removing discontinuities (rationalization)

Sal finds the value the function f(x)=(√(x+4)-3)/(x-5) should have at x=5 so it's continuous at that point. Created by Sal Khan.

## Want to join the conversation?

• I didn't understand the last part
how is c=1/6??
because 1/6 is the limit as 'x' approaches 5 but not x=5 • I understand most everything in the video, but have one question that really plagues me whenever I am doing math (Sal seemed to touch on this indirectly). How do you know when to take the negative/positive square root of a number? In this case, Sal took +3 because it was the "principal root" - what is that and why did he take it?

If someone could please explain this to me in simple terms, that would be greatly appreciated. • The general understanding is that if it is you who introduces a square root, you should specify which you mean with it being understood to be the principal square root unless you write `−` or `±` to indicate otherwise.

However, if you are the one performing the square root from another function, then you would normally use both square roots unless you have reason to do otherwise.
thus:
For `x² = 3` you would use both square roots in finding `x = ± √3` because it was you who introduced the square root.
But for `3 - √x = 42` you would only use the negative square root because the person giving the problem made it clear which was meant.

In the case presented in the video there is no sign in front of the square root, we just have the square root presented to us with no indication to consider the negative square root. There, we assume this is the principal square by default unless some circumstance requires us to do otherwise.

Unfortunately, sometimes the situation is ambiguous.
• OK I don't wona sound stupid but what is the point of limits if function is continuous ? • Let's say that you want to evaluate the limit of f(x) as x approaches v. If f is continuous around x=v and you can easily evaluate f(v), then the limit is just f(v) and there isn't much you have to do.

In this case, v is 5. However, we don't know what f(5) is so even though the limit of f(x) as x approaches 5 is just f(5), we still need to find f(5). Luckily, we know that f(x) for x does not equal v is [√(x+4)-3]/(x-5), so we can find the limit of that as x approaches 5 to find f(5).

Thus, limits of continuous functions can help us "fill in the gaps" of values we don't know.

I hope this helps!
• At , why don't the (x-5) terms just cancel? • Could you also just plug in 4.99999 and 5.00001 into your calculator and see what value that approaches? • You certainly can do that but you may not always be able to tell what it is actually approaching unless you know the value already. So if I had a function that approaches 2.64575131106 as x approaches some number, it can also be misleading because you may be approaching something, like say 1360149/514088 or just the square root of 7. Even though these numbers are very close, you won't get the precise answer unless you actually solve the problem.
• I don't get it. He just said that it's the limit when x is NOT 5, because 5-5 = 0. So how can this be continuous if f(5) is undefined? • This is a piecewise function, which means that the function behaves differently at different x values. Everywhere where x isn't equal to 5, the function is the one that Sal worked with during the video. When x is equal to 5, the function is just equal to 1/6, so f(5) is defined. The limit of the more complicated function is 1/6 when x approaches 5, and since the limit of f(5) equals the definition of f(5), it is continuous. (Rather, you're trying to find the value of c such that the function is continuous, which in this case is 1/6.)
• I have a general question about continuity. If the function has an asymptote and you are asked to discuss the continuity how would you go about doing that? • In the equation provided, the answer becomes simple cause the numerator and denominator eventually have a (x-5) in it. But what if the equation is (sqrt(x+1)-3)/(x-5)? I can't seem to do it in the same method. • What is the difference between "undefined" and "indeterminate"? I understand that any non-zero number divided by zero is undefined and zero divided by zero is indeterminate. But my question is - how does it matter? Why do we just leave it as it is when it is undefined and why do we simplify the equation when it is in indeterminate form?
(1 vote) • This video centers around the question,

Let f be the function given by
f(x) = { (sqrt(x+4)-3) / (x-5) if x=/=5
c if x=5}
If x is continuous at x = 5, then what is the value of c?

Isn't that just the same thing as asking "what is the limit of f as x approaches c"?
Why use unfamiliar wording like "If f is continuous at x = 5, then what is the value of c"?

Some of my peers have spent the entire video just trying to understand what the latter question was asking instead of listening to Sal. I understand this topic centers around continuity, so is there a particular reason behind the wording (as opposed to simply asking for the limit)? • To understand the question, it helps to understand the graph or piecewise function. The function is obviously undefined at x=5 (hence x cannot = 5), then second piece of the function says that if x=5, then y = c (an arbitrary point that isn't on the first function). If you visualize this, it is a function with a removable discontinuity (f is not continuous at x=5).
What the question asks, if f is continuous at x=5 (meaning move the point (5,c) into the hole/gap of the function where it was undefined at x=5) what would the numerical value for c be?

If c is going to be in the place of that hole/gap then c is going to be the limit as x->5, which is what Sal is saying in the video. c is a y-value so it's not x approaches c per your question, but x approaches 5.

You are correct it's about continuity, hence the question word this way, to make sure you understand what continuity mean. If they had asked, "what is the limit as x approaches 5 of f(x)?" then it's about limit, not continuity anymore.

The definition of continuity is "lim_x->c f(x) = f(c)"

The main focus of this question is to understand the above definition.

Hope that helps.
(1 vote)

## Video transcript

Let f be the function given by f of x is equal to the square root of x plus 4 minus 3 over x minus 5. If x does not equal 5, and it's equal to c if x equals 5. Then say, if f is continuous at x equals 5, what is the value of c? So if we know that f is continuous at x equals 5, that means that the limit as x approaches 5 of f of x is equal to f of 5. This is the definition of continuity. And they tell us that f of 5, when x equals 5, the value of the function is equal to c. So this must be equal to c. So what we really need to do is figure out what the limit of f of x as x approaches 5 actually is. Now, if we just try to substitute 5 into the expression right up here, in the numerator you have 5 plus 4 is 9. The square root of that is positive 3, the principal root is positive 3. 3 minus 3 is 0. So you get a 0 in the numerator. And then you get 5 minus 5 in the denominator, so you get 0 in the denominator. So you get this indeterminate form of 0/0. And in the future, we will see that we do have a tool that allows us, or gives us an option to attempt to find the limits when we get this indeterminate form. It's called L'Hopital's rule. But we can actually tackle this with a little bit of fancy algebra. And to do that, I'm going to try to get this radical out of the numerator. So let's rewrite it. So we have the square root of x plus 4 minus 3 over x minus 5. And any time you see a radical plus or minus something else, to get rid of the radical, what you can do is multiply by the radical-- or, if you have a radical minus 3, you multiply by the radical plus 3. So in this situation, you just multiply the numerator by square root of x plus 4 plus 3 over the square root of x plus 4 plus 3. We obviously have to multiply the numerator and the denominator by the same thing so that we actually don't change the value of the expression. If this right over here had a plus 3, then we would do a minus 3 here. This is a technique that we learn in algebra, or sometimes in pre-calculus class, to rationalize usually denominators, but to rationalize numerators or denominators. It's also a very similar technique that we use often times to get rid of complex numbers, usually in denominators. But if you multiply this out-- and I encourage you to do it-- you notice this has the pattern that you learned in algebra class. It's a difference of squares. Something minus something times something plus something. So the first term is going to be the first something squared. So square root of x plus 4 squared is x plus 4. And the second term is going to be the second something, or you're going to subtract the second something squared. So you're going to have minus 3 squared, so minus 9. And in the denominator, you're of course going to have x minus 5 times the square root of x plus 4 plus 3. And so this has-- I guess you could say simplified to, although it's not arguably any simpler. But at least we have gotten our radical. We're really just playing around with it algebraically to see if we can then substitute x equals 5 or if we can somehow simplify it to figure out what the limit is. And when you simplify the numerator up here, you get x plus 4 minus 9. Well, that's x minus 5 over x minus 5 times the square root of x plus 4 plus 3. And now it pops out at you, both the numerator and the denominator are now divisible by x minus 5. So you can have a completely identical expression if you say that this is the same thing. You can divide the numerator and the denominator by x minus 5 if you assume x does not equal 5. So this is going to be the same thing as 1 over square root of x plus 4 plus 3 for x does not equal 5. Which is fine, because in the first part of this function definition, this is the case for x does not equal 5. So we could actually replace this-- and this is a simpler expression-- with 1 over square root of x plus 4 plus 3. And so now when we take the limit as x approaches 5, we're going to get closer and closer to five. We're going to get x values closer and closer to 5, but not quite at 5. We can use this expression right over here. So the limit of f of x as x approaches 5 is going to be the same thing as the limit of 1 over the square root of x plus 4 plus 3 as x approaches 5. And now we can substitute a 5 in here. It's going to be 1 over 5 plus 4 is 9, principal root of that is 3. 3 plus 3 is 6. So if c is equal to 1/6, then the limit of our function as x approaches 5 is going to be equal to f of 5. And we are continuous at x equals 5. So it's 1/6.