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

Lesson 7: Determining limits using algebraic manipulation# Limits by rationalizing

In this video, we explore how to find the limit of a function as x approaches -1. The function is (x+1)/(√(x+5)-2). To tackle the indeterminate form 0/0, we "rationalize the denominator" by multiplying the numerator and denominator by the conjugate of the denominator. This simplifies the expression, allowing us to evaluate the limit.

## Want to join the conversation?

- At6:09, Sal introduces the concept to use another function to solve for the limit of g(x). Wouldn't it just be easier to just plug in the limit as x to the simplified version of g(x) without having to use f(x)?(49 votes)
- The point he is making here is perhaps a bit unclear. He's not actually using another function to solve for the limit, per se. He's showing mathematical proofs - that if there are two functions, which are identical except for a discontinuity at a single point, then their limits will be the same, so you can use the limit from the continuous function (which you can get from just plugging in to the equation) for the discontinuous function. So that whole f(x) bit is the mathematical proof that lets you use the technique you just mentioned - just plugging the number into the simplified expression - to get the limit.

I probably would have started with, "If that discontinuity wasn't there, what would the value be? Well, we'd have 'the square root of x plus five, plus two', which at x = -1 would be 4. The discontinuity is there, so the value isn't 4, but we can see that it would be 4 if the hole wasn't there, so the limit is 4." And then given the written out proof that he gives here to show how that is mathematically valid rather than just graphically intuitive.(172 votes)

- Oh! That makes sense! Anything
**other**than 0 divided by 0 is undefined, but 0/0 is indeterminate, which allows you to manipulate it into other forms. I never understood whether or not I'd be justified in manipulating an expression like this into a nicer form. Apparently, it's justified when the numerator is zero. I always see the algebraic manipulations immediately when I see a problem like this, but I always worry it might not be justified...

My question: why IS such manipulation permissible when the numerator is zero? And relatedly, why is this NOT justified when the numerator is non-zero?

Thank you very much for your help!(44 votes)- This may not be an answer for all cases, but for rational functions (functions with polynomial numerators and denominators) a non-zero number divided by zero indicates the presence of a vertical asymptote in the graph. In most cases, we say the limit of a function does not exist as it approaches an asymptote, so there's no manipulation you could do to find a limit. However (still for rational functions) zero divided by zero indicates the presence of a removable point discontinuity, or "hole." In these cases, though the function does not have a value at that point, it does have a limit, so manipulating it could allow you to find that limit. It is possible this is true of other situations that yield division by zero, but I don't know enough to really say.

This may not be the most valid math (or maybe it is!) but here's another way to think of it: The reason a non-zero number divided by zero is undefined is there is absolutely nothing it could equal under the existing rules and definitions of numbers and math. Turn around an equation such as 2/0 = x and it becomes 0x = 2. There is no number you can multiply by zero and get two! In terms of limits, there is none to be found. But the reason zero divided by zero is undefined is that it could theoretically be any number. Turn around 0/0 = x and it becomes 0x = 0. Anything times zero is zero! In terms of limits, there is a limit there to be found. It's obscured by the 0/0, but some manipulation could reveal it.(77 votes)

- if we just look at " sqrt(x+5) -2 " as the denominator without rationalizing, we know it cannot equal zero. So we solve for zero:

"sqrt(x+5) -2 = 0" --> x = -1

alright, what if we test this? sqrt(-1+5) -2 = 0 --> sqrt(4)-2 =0

BUT, sqrt(4) could equal 2 OR -2.

what's up with this -2?(27 votes)- As Sal said at1:20, we take principal root, meaning we only take the positive number.(50 votes)

- Why to we want principle square roots only?(10 votes)
- For a function to be a function, every input must have a single output. Since each number has two square roots, we need a convention for which one to use. The convention is that the square root sign by itself indicates only the principal (in this case positive) square root. A negative sign in front of the radical would indicate the negative root, and a plus-or-minus sign would indicate both.(15 votes)

- Instead of going through the rationalizing process to find the limit of g(x), is it reasonable to just replace x with something really close to -1 (like -0.99999) so that g(x) doesn't equal to 0/0 and then solving for g(x) using that really close value?(11 votes)
- That often works for these problems, because we can expect the answer to be a clean integer or simple rational number. But in general, we can't quite tell what number we're approaching with this method. It could appear to be approaching 0, but the limit is actually some small positive number, for example.(8 votes)

- Is there any way for us to say if a function is continuous or not without knowing the limit or graphing it? How does Sal say with such certainty that f(x), for example, is continuous?(7 votes)
- We know several things about continuous functions:

1. All polynomials are continuous (including constant and linear functions)

2. The square root function is continuous

3. The sum of continuous functions is continuous

4. The quotient of continuous functions is continuous whenever it's defined

5. The composition of continuous functions is continuous whenever it's defined

We can prove all these things from the epsilon-delta definition of continuity.

Assembling these facts, we can see that f(x) is continuous wherever it's defined (it's undefined, and therefore not continuous, at x= -1).(11 votes)

- But how did sal know that the conjugate would work? Are there situations where it wouldn't work and if so how do I know what these are?(8 votes)
- For cases where you have square-roots on the denominator, it will always work. There's a whole bunch of methods you can use if that doesn't work, but they'd take too long to go through here.(6 votes)

- i love videos that end with "if it confuses you, ignore it".(10 votes)
- What if the square root expression is in the numerator?(5 votes)
- Rationalizing the numerator is also a valid strategy. The purpose of rationalizing is to change the expression into a form that might be easier to work with. From there, you might be able to cancel and simplify until you can use direct substitution. These exercises have removable discontinuities, so any simplification will not change the value of the limit.(6 votes)

- at4:09, instead of multiplying it by sqrt(x+5) +2/sqrt(x+5) +2 , cant we just multiply it by sqrt(x+5) -2/sqrt(x+5) -2 ?(5 votes)
- You could do so, but in this case it wouldn't help. Both numerator and denominator would be more complex than before and you would still get 0/0. Furthermore, it wouldn't provide a matched pair of factors in the numerator and denominator, which are necessary for "removing" the discontinuity.

sqrt(x + 5) + 2 is specifically useful because, as the conjugate of sqrt(x + 5) - 2 , it removes the radical in the denominator, and in this case it's the only thing you can multiply by to get an (x + 1) in the denominator, which you can use to "remove" the discontinuity.(6 votes)

## Video transcript

- [Voiceover] Let's see
if we can find the limit as x approaches negative 1 of x plus 1 over the square root of x plus 5 minus 2. So, our first reaction might just be, okay, well let's just
use our limit properties a little bit, this is going to be the
same thing as the limit as x approaches negative 1 of x plus 1 over, over the limit, the limit as x approaches negative 1 of square root of x plus 5 minus 2. And then we could say, all right, this thing up here, x plus 1, if this is, if we think about the
graph y equals x plus 1, it's continuous everywhere, especially at x equals negative 1, and
so to evaluate this limit, we just have to evaluate
this expression at x equals negative 1, so this numerator's just going
to evaluate to negative 1 plus 1. And then our denominator, square root of x plus 5 minus
2 isn't continuous everywhere but it is continuous
at x equals negative 1 and so we can do the same thing. We can just substitute negative 1 for x, so this is going to be the
square root of negative 1 plus 5 minus 2. Now, what does this evaluate to? Well, in the numerator we get a zero, and in the denominator,
negative 1 plus 5 is 4, take the principle root is 2, minus 2, we get zero again, so we get, we got zero over zero. Now, when you see that, you
might be tempted to give up. You say, oh, look, there's
a zero in the denominator, maybe this limit doesn't exist, maybe I'm done here, what do I do? And if this was non-zero
up here in the numerator, if you're taking a non-zero
value and dividing it by zero, that is undefined and your limit would not exist. But when you have zero over zero, this is indeterminate form, it doesn't mean necessarily
that your limit does not exist, and as we'll see in this
video and many future ones, there are tools at our
disposal to address this, and we will look at one of them. Now, the tool that we're going to look at is is there another way of
rewriting this expression so we can evaluate its limit without getting the zero over zero? Well, let's just rewrite, let's just take this and give it, so let's take this thing right over here, and let's say this is g of x, so essentially what we're
trying to do is find the limit of g of x as x approaches negative 1, so we can write g of x is equal to x plus 1 and the only reason
I'm defining it as g of x is just to be able to think of
it more clearly as a function and manipulate the function and then think about similar functions, over x plus 5 minus 2, or x plus 1 over the square
root of x plus 5 minus 2. Now, the technique we're
going to use is when you get this indeterminate form and
if you have a square root in either the numerator
or the denominator, it might help to get
rid of that square root and this is often called
rationalizing the expression. In this case, you have a
square root in the denominator, so it would be rationalizing the denominator, and so, this would be, the way we would do it is we would be leveraging
our knowledge of difference of squares. We know, we know that a plus b times a minus b is equal to a squared minus b squared, you learned that in algebra a little while ago, or if we had the square root of a plus b and we were to multiply that
times the square root of a minus b, well that would be
the square root of a squared which is just going to be a, minus b squared, so we can just leverage these ideas to get rid of this radical down here. The way we're going to do it is we're going to multiply the
numerator and the denominator by the square root of x plus 5 plus 2, right? We have the minus 2 so we multiply it times the plus 2, so let's do that. So we have square root of x plus 5 plus 2 and we're going to multiply
the numerator times the same thing, 'cause we
don't want to change the value of the expression. This is 1. So, if we take the expression
divided by the same expression it's going to be 1, so this is, so square root of x plus 5 plus 2 and so this is going to be equal to, this is going to be equal to x plus 1 times the square root, times the square root of x plus 5 plus 2 and then the denominator is going to be, well, it's going to be x, the square root of x plus 5 squared which would be just x plus 5 and then minus 2 squared, minus 4, and so this down here
simplifies to x plus 5 minus 4 is just x plus 1 so this is just, this is just x plus 1 and it probably jumps out at
you that both the numerator and the denominator
have an x plus 1 in it, so maybe we can simplify, so we can simplify by just say, well, g of x is equal to
the square root of x plus 5 plus 2. Now, some of you might be
feeling a little off here, and you would be correct. Your spider senses would be, is this, is this definitely the same thing as what we originally had
before we cancelled out the x plus 1s? And the answer is the way I just wrote it is not the exact same thing. It is the exact same
thing everywhere except at x equals negative 1. This thing right over here is defined at x equals negative 1. This thing right over here is not defined at x equals negative 1, and g of x was not, was not, so g of x right over here, you don't get a good result
when you try x equals negative 1 and so in order for this
to truly be the same thing as g of x, the same function, we have to say for x not equal to negative 1. Now, this is a simplified
version of g of x. It is the same thing. For any input x, that g of x is defined, this
is going to give you the same output, and this is the
exact same domain now, now that we've put this constraint in, as g of x. Now you might say, okay,
well how does this help us? Because we want to find the
limit as x approaches negative 1 and even here, I had to put
this little constraint here that x cannot be equal to negative 1. How do we think about this limit? Well, lucky for us, we know, lucky for us we know that if we just take
another function, f of x, if we say f of x is equal to
the square root of x plus 5 plus 2, well then we know that f
of x is equal to g of x for all x not equal to negative 1 because f of x does not
have that constraint. And we know if this is true of two, if this is true of two functions, then the limit as x approaches, the limit, let me write this down, since we know this, because of this, we know that the limit of f of x as x approaches negative
1 is going to be equal to the limit of g of x as x approaches negative 1, and this of course is what
we want to figure out, what was the beginning of the problem, but we can now use f of x here, because only at x equals negative 1 that they are not the same, and if you were to graph g of x it just has a, it has a point discontinuity, or removable discont-- or, I should say, yeah, a point discontinuity right over here at x equals negative 1, and so what is the limit? And we are in the home stretch now. What is the limit of f of x? Well, we could say the limit
of the square root of x plus 5 plus 2 as x approaches negative 1, well, this expression is continuous. Or, this function is continuous
at x equals negative 1 so we can just evaluate
it at x equals negative 1, so this is going to be the
square root of negative 1 plus 5 plus 2, so this is 4 square root, principle root of 4 is 2, 2 plus 2 is equal to 4. So since the limit of f of
x as x approaches negative 1 is 4, the limit of g of x as x approaches negative 1 is also 4, and if this little, this little, I guess you could say leap
that I just made over here doesn't make sense to you, think about it, think about it visually. Think about it visually. So if this is my y-axis and this is my x-axis, g of x looked something like this. G of x, g of x, let me draw it, g of x looked something, something like this, and it had a gap at negative 1, so it had a gap right over there, while f of x, f of x would have the same graph except it wouldn't have, it wouldn't have the gap, and so if you're trying to find the limit, it seems completely reasonable, well let's just use f
of x and evaluate what f of x would be to kind of fill that gap at x equals negative 1, so hopefully this graphical
version helps a little bit or if it confuses you, ignore it.