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