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• - [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.

• 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