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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
• he should have asked what is the value of f(c) ,,because the value of is already given .i.e. 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?
• 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.
• 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?
• A vertical asymptote is a discontinuity on a function, so you can say that the function is discontinuous at the vertical asymptote.
• 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.
• 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.
• 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)
• At he 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!