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## Calculus, all content (2017 edition)

### Course: Calculus, all content (2017 edition) > Unit 1

Lesson 17: Removable discontinuities# Removing discontinuities (rationalization)

AP.CALC:

LIM‑2 (EU)

, LIM‑2.C (LO)

, LIM‑2.C.1 (EK)

, LIM‑2.C.2 (EK)

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(11 votes)- he should have asked what is the value of f(c) ,,because the value of is already given .i.e. 5(4 votes)

- 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.(6 votes)- 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.(22 votes)

- OK I don't wona sound stupid but what is the point of limits if function is continuous ?(9 votes)
- 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!(7 votes)

- At4:07, why don't the (x-5) terms just cancel?(6 votes)
- It does, and that's what he writes.

x is not allowed to be 5 in the original equation, so that constraint must be kept after cancelling the x-5 term.(3 votes)

- Could you also just plug in 4.99999 and 5.00001 into your calculator and see what value that approaches?(3 votes)
- 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.(6 votes)

- 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?(3 votes)
- 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.)(5 votes)

- 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?(3 votes)
- A vertical asymptote is a discontinuity on a function, so you can say that the function is discontinuous at the vertical asymptote.(2 votes)

- 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.(2 votes)
- Well, in that case the numerator doesn't approach zero as x approaches 5.

The numerator approaches √6 - 3 as the denominator approaches zero, so the function heads off towards negative infinity.(2 votes)

- 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)
- Undefined means that there is no answer. Indeterminate means that there is an answer, but we must rearrange the equation to find it. (this is not a technical explanation, but it should do well enough)(4 votes)

- At1:07he says this can be solved with L'Hopital's rule as well. I got curious and was able to solve his problem with the rule, but was not able to solve mine with it as I got an answer of 2sqrt(-16). Can anyone solve the problem in the link using L'Hopital's rule?

https://goo.gl/Uyqc8x(1 vote)- lim x-->3 of (x-3)/[sqrt(x+13) - 4]

= lim x-->3 of [d/dx of (x-3)] / {d/dx of [sqrt(x+13) - 4]}, since the original limit has a 0/0 form

= lim x-->3 of [d/dx of (x-3)] / {d/dx of [(x+13)^(1/2) - 4]}

= lim x-->3 of 1 / [(1/2)(x+13)^(-1/2)]

= lim x-->3 of 2(x+13)^(1/2)

= lim x-->3 of 2sqrt(x+13)

= 2sqrt(16)

= 8.

Have a blessed, wonderful day!(3 votes)

## 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.