Limits by rationalizing | Limits and continuity | AP Calculus AB | Khan Academy

Let's see if we can find the limit as x approaches negative one of ( \frac{x + 1}{\sqrt{x + 5} - 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 one of ( \frac{x + 1}{\sqrt{x + 5} - 2} ). And then we could say, all right, this thing up here ( x + 1 ), if we think about the graph ( y = x + 1 ), it's continuous everywhere, especially at ( x = -1 ).

And so to evaluate this limit, we just have to evaluate this expression at ( x = -1 ). So this numerator is just going to evaluate to ( -1 + 1 ). And then our denominator ( \sqrt{x + 5} - 2 ) isn't continuous everywhere, but it is continuous at ( x = -1 ). And so we can do the same thing; we can just substitute ( -1 ) for ( x ).

So this is going to be ( \sqrt{-1 + 5} - 2 ). Now, what does this evaluate to? Well, in the numerator we get a zero, and in the denominator, ( -1 + 5 ) is 4. Taking the principal root is 2, and minus 2 gives us zero again. So we got ( \frac{0}{0} ).

Now, when you see that, you might be tempted to give up. You see, oh look, there's a 0 in the denominator; maybe this limit doesn't exist, maybe I'm done here. What do I do? And if this was a non-zero value 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 ( \frac{0}{0} ), this is indeterminate form, and it doesn't mean necessarily that your limit does not exist.

As we'll see in this video and in 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 that we can evaluate its limit without getting the ( \frac{0}{0} )?

Well, let's just rewrite this; let me give it. So let's take this thing right over here and say this is ( g(x) ). So essentially, what we're trying to do is find the limit of ( g(x) ) as ( x ) approaches negative one. So we could write ( g(x) = x + 1 ).

And the whole reason why I'm defining it as ( g(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 ( \frac{x + 1}{\sqrt{x + 5} - 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 denominator, it might help to get rid of that square root.

This is often called rationalizing expression. In this case, you have a square root in the denominator, so it would be rationalizing the denominator. The way we're going to do it is by leveraging our knowledge of the difference of squares. We know that ( (a + b)(a - b) = a^2 - b^2 ).

You learned that in algebra a little while ago. Or if we had the square root of ( a + b ), and we were to multiply that times the square root of ( a - b ), well, that'd be the square root of ( a^2 ) which is just ( a - b^2 ). 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 denominator by ( \sqrt{x + 5} + 2 ). Right? We have the minus 2, so we multiply it times the plus 2. So let's do that. We have ( \sqrt{x + 5} + 2 ), and we're going to multiply the numerator times the same thing because 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, ( \sqrt{x + 5} + 2 ). Thus, this is going to be equal to ( (x + 1)(\sqrt{x + 5} + 2) ), and then the denominator is going to be well, it's going to be ( (\sqrt{x + 5})^2 - 2^2 ).

And so this down here simplifies to ( x + 5 - 4 ), which is just ( x + 1 ). So this is just ( x + 1 ). It probably jumps out at you that both the numerator and the denominator have an ( x + 1 ) in it, so maybe we can simplify.

So we could simplify and just say, well, ( g(x) = \sqrt{x + 5} + 2 ). Now, some of you might be feeling a little off here, and you would be correct. Your spider senses would be saying, is this definitely the same thing as what we originally had before we cancelled out the ( x + 1 )'s?

The answer is, the way I just wrote it, it is not the exact same thing; it is the exact same thing everywhere except at ( x = -1 ). This thing right over here is defined at ( x = -1 ); this thing right over here is not defined at ( x = -1 ), and ( g(x) ) was not defined at ( x = -1 ).

So in order for this to truly be the same thing as ( g(x) ), the same function, we have to say for ( x \neq -1 ). Now, this is a simplified version of ( g(x) ); it is the same thing for any input ( x ) that ( g(x) ) is defined. This is going to give you the same output, and this has the exact same domain.

Now that we've put this constraint in as ( g(x) ), now you might say, okay, well how does this help us? Because we want to find the limit as ( x ) approaches negative one, and even here I had to put this little constraint here that ( x ) cannot be equal to negative one.

How do we think about this limit? Well, lucky for us, we know that if we just take another function ( f(x) ), if we say ( f(x) = \sqrt{x + 5} + 2 ), then we know that ( f(x) = g(x) ) for all ( x \neq -1 ) because ( f(x) ) does not have that constraint.

We know if this is true of two functions, then the limit as ( x ) approaches negative one of ( f(x) ) is going to be equal to the limit of ( g(x) ) as ( x ) approaches negative one. This, of course, is what we want to figure out was the beginning of the problem, but we can now use ( f(x) ) here because only at ( x = -1 ) that they are not the same.

If you were to graph ( g(x) ), it just has a point discontinuity, or removable discontinuity, or I should just say a point discontinuity right over here at ( x = -1 ).

And so what is the limit? And we are in the home stretch now. What is the limit of ( f(x) ) or we could say the limit of ( \sqrt{x + 5} + 2 ) as ( x ) approaches negative one? Well, this expression is continuous; this function is continuous at ( x = -1 ), so we can just evaluate it at ( x = -1 ).

So this is going to be ( \sqrt{-1 + 5} + 2 ). So this is 4; the principal root of 4 is 2, and 2 plus 2 is equal to 4. So since the limit of ( f(x) ) as ( x ) approaches negative one is 4, the limit of ( g(x) ) as ( x ) approaches negative one is also 4.

If this little leap that I just made right over here doesn't make sense to you, think about it visually. So if this is my y-axis and this is my x-axis, ( g(x) ) looked something like this. Let me draw it; ( g(x) ) looked something like this, and it had a gap at negative one.

So it had a gap right over there, while ( f(x) ) would have the same graph except it wouldn't have the gap. So if you're trying to find the limit, it seems completely reasonable: let's just use ( f(x) ) and evaluate what ( f(x) ) would be to kind of fill that gap at ( x = -1 ).

So hopefully, this graphical version helps a little bit, or if it confuses you, ignore it.