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## Calculus, all content (2017 edition)

### Unit 1: Lesson 17

Removable discontinuities

# Removing discontinuities (factoring)

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)=(6x²+18x+12)/(x²-4) should have at x=-2 so it's continuous at that point. Created by Sal Khan.

## Want to join the conversation?

• At , I understand why x cannot equal 2, but why can it not equal -2?
• Because he cancelled out an (x + 2) factor earlier.
-2 was not allowed in the original equation, so you can't allow it in your simplified equation.
• What if we want to find what value should be assigned to f(2) to make f(x) continuous at that point. Is that possible?
• I haven't learnt much, but this is what I understand:

Notice that, in the original equation, inputting x = -2 will evaluate to 0/0. This is meaningless, and if you plot the equation on a graphing software you'll see the graph on either side of this point is quite smooth, apart from just one hole here at x=-2. I don't know how to show graphs here, but here are a few values:
(-1.9,1.38)
(-2.0,?)
(-2.1,1.61)
It would seem to "make sense" to assign a number to that question mark in between these,
By making x really close to -2 you can get the corresponding y value to be as close as 1.5 as your efforts in making x close to -2. That's what the limit means.
We say the limit as x approaches -2 of f(x) = 1.5

However, when you input x=+2 to the original equation, you get 72/0 which shows that the graph is curving up towards infinity here at x=+2. So it's not possible to make the function continuous here.

Some are talking about some sort of hospital 🏥 rule, I haven't learnt that yet so sorry if I'm wrong
• So, by defining the point to make the function continuous, aren't you also finding the limit of the function at the discontinuity? Would it be correct to say of the original function: lim f(x) as x-->-2 = 3/2? There really is no defined value for x=+/- 2, correct?
• You are correct , in the original function, call it f(x), is not defined at x = +/-2. Sal finds the limit of f(x) as x approaches -2, by finding a new function, let's call it g(x), that is the same as f(x) EXCEPT that g(x) is defined at x = -2. So it is simple to find the limit of the g(x) at x = -2. Since f(x) and g(x) differ only at x = -2, where f(x) has no value, and g(x) has a value, that value is the limit of both functions as x approaches -2. Now we know the limit of f(x) as x approaches -2. So we can create a completely new function, call it h(x), that very specifically says what happens when x = -2. Something like:

h(x) = whatever was in the original function for x not equal to -2
3/2 for x = -2

Now the limit of h(x) as x approaches -2 is the same as the value of h(x) when x = 2, thus satisfying the definition of continuous, so h(x) is continuous at x = -2.
• I don't understand how this makes the function continuous. X= -2 is one of the vertical asymptotes of this function. Can the function really reach 3/2? What does "assign a value" mean?
• Another way to look at this is that the value of the function at x = -2 is only ambiguous because we are dividing by 0 when x = -2. If you simply take the limit of the function as x --> -2, the limit = 3/2. What is being done here is assigning the limiting value to the point that is ambiguous because we have zero divided by zero when x = - 2. There are no asymptotes at x = -2. There are asmyptotes at x = +2 and it is not possible to assign a value here to make the function continuous.
• How can he first say that f(x) = 3/2 only if x =/ -2, but then use that to say that f(-2) = 3/2 ?
• Because the original question was asking him to fill in the "removable" discontinuity at f(-2), which he did by figuring out the limit of f(x) when approaching -2 with algebra. If you were to plug in numbers that were infinitely close to -2 into f(x) you would come up with the same answer. The new piecewise equation fills in that chunk so that the graph is continuous at f(-2).
• Division by 0 is not defined? This is something I've never understood as to why it is not defined. Even when I was a kid and teacher told me that. I tried dividing 1 by a very very small number, the smaller the number was, when 1 was divided by it, it would become bigger. And as I started into the limits section of KhanAcademy this made even more and more sense to me. It shouldn't be undefined when it's clearly approaching infinity. Dividing by zero basically is infinity.
• The reason you cannot define division by zero is that it gives you inconsistent answers. You can, in fact, use division by zero to make it look like any number is equal to any other number.

A limit is not the same as actually using the value that you are approaching.
Also, infinity is not a number.
• having trouble with using this method for f(x)= x+1/(xe^x+e^x), x not equal to -1, find f(-1)
• Take e^x common in the denominator and cancel x+1 as x tends to -1. I suggest reviewing relevant factorization methods as they become very important in limits and beyond.
• Okay, I'm doing the Continuity exercises. Since there seems to be no place for me to ask questions about them, I'll try my luck here.

It gives me a function with 3 parts, for x < 0, 0 <= x <= 2, x>2. It asks me for the values of A and B to make F continuous. Why when it starts solving it, it tries to see if f(x) is continuous at x=0 and x=2. Why do we care only about those specific numbers? Is it because they are the only numbers in which the function has an abnormality?
• If a function is defined only at one point,is it continuous there ?help me!
• Yes, a function defined at only one point, is continuous at that point.

This is perhaps a bit more abstract than what you have in mind, but the property of being continuous at the only point of definition may be somewhat extended as follows:

Suppose `U` is any non-empty subset of the real numbers, `R`. We say that a point `u` in `U` is an isolated point of `U` if and only if there exists a real number `r > 0` such that the open interval `(u - r, u + r)` contains no point of `U` other than `u`; in other words, `u` is an isolated point of `U` if there exists some open interval about `u` which contains no other points of `U`.

Suppose now that `ƒ: U → R` is a real-valued function defined on `U`. If `u` is an isolated point of `U`, then `ƒ` is continuous at `u` (this may be proved). As a corollary, if every point of `U` is an isolated point of `U`, then `ƒ` is continuous on `U`.

As a result, we arrive at the following (counterintuitive?) result: since any sequence of real numbers is a real-valued function defined on the set of natural numbers, and since every natural number is an isolated point of the set of natural numbers (can you prove this? Hint: take `r = 1/2` in the definition above), every sequence of numbers is a continuous function! (This fact is mainly a result of extreme cases of the definition of continuity.)
• At , what is the definition of 'constrain'?
(1 vote)
• A user, by the name of dashpointdash, explained this in the video "Limit and function defined at point of discontinuity". I am about to paste, word by word, the users quote as it was appreciated by other users:

A constraint is a condition, which has to be fullfilled by the object you're observing.
e.g. consider the description of primes in the set of numbers like this one
"Primes are only numbers divisible by one and itself".

My explanation: A constraint is a condition. A constraint can be conveyed by using a word like 'however'....e.g look at this example "You can play outside. However, you have to come back by 6". The constraint in this case is the command of coming back by 6.

Whichever explanation you prefer, I hope you understood the meaning of the word constraint.