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My cousin Nadia is taking a summer calculus course and

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she called me last night.

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And she had some limit problems and they were

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excellent problem.

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So I thought it was worthwhile to make videos on the problems

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that she had to figure out.

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Anyway, so let's do them.

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If I remember them correctly.

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I'm doing this on memory.

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So I hope the answers work out like they did last night when

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we were going over them on the phone.

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So if I remember correctly, the first one was the limit

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as z approaches 2 of x.

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Oh no.

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Not that-- There's no x.

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It's a z.

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Of z squared plus 2z minus 8.

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All of that over z to the fourth minus 16.

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So the first thing when you want to do when you try a limit

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problem is, well you just try to substitute the value into--

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Maybe there's no problem when z equals 2 and you just evaluate

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the function at z equals 2.

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If it's a continuous function, then the limit as z

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approaches 2 is going to be the function at 2.

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But you immediately see a problem if you put 2 right

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here-- 2 to the fourth minus 16 --and you get a 0 in the

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denominator, which is undefined, so we have to

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figure out some way to get around that.

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And nine times out of 10, when you see a problem like this,

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the solution, because both the numerator and the denominator

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look factorable, is to factor the numerator and

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the denominator.

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So this is equal to the limit as z approaches 2.

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What's the numerator factored?

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Let's see.

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Something when you add two numbers, you get positive

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2, and you multiply them, you get minus 8.

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So it's probably plus 4 and minus 2, right?

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So this is going to be x plus-- No.

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Not an x.

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I'm so conditioned.

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I'm like a dog.

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Oh.

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I can't even undo it anymore.

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Anyway, well z-- Actually let me erase that.

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I don't want to be messy.

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Let me erase.

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I tried to undo it and it doesn't remember everything.

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So it's z plus 4 times z minus 2.

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That's just factoring a quadratic.

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And what is this?

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This has the form a squared minus b squared, right?

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So that's going to be-- But a squared-- If this is a squared

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minus b squared, a would be z squared, right?

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So it's z squared plus 4 times z squared minus 4.

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And then, of course, this also has the form a squared

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plus b squared, right?

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So this will further factor into z plus 2 times z minus 2.

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Well, our factoring paid off.

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We see a term in the numerator and the denominator

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that are the same.

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And not only are they the same, but they-- This is

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the term that was giving us the problems, right?

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Because you put a 2 in here, you got a 0.

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So let's assume that we're not evaluating it at z equals 2.

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And so, for all other values, we can divide those

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because those are going to be the same values.

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And then what are we left with?

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This is equal to the limit-- and I've change colors

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arbitrarily --as z approaches 2 of z plus 4 over z squared

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plus 4 times z plus 2.

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Which is equal what?

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That's equal to 6 over-- what's 2 squared plus 4?

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Is 8.

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And then 2 plus 2 is 4.

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6 over 12 is equal to 1/2.

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There you go.

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I thought that was a pretty interesting problem.

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Let's do another one.

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And this one I found even more interesting.

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She gave me-- It's really testing my memory to see

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if I can-- But I remember the gist of the problem.

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So I might not get the exact numbers she'd given me, but

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hopefully I get the exact properties.

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So it's the limit as x approaches infinity of the

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square root of x squared plus 4x plus 1 minus x.

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So when you look at this you're like well let's see.

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What's happening here?

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Let's see this.

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As you go to infinity, this term will get really big.

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But then we're taking the square root of it.

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And it seems like this term would overpower this term.

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And then, so maybe this kind of converges to x, but then

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we would subtract an x.

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So maybe it approaches 0.

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So that's, at least, that was my first intuition

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when I spoke to her.

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But as we will see, the intuition here is wrong.

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And really to do this you have to know a little

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bit of a trick.

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And this trick pays off a lot whenever you see something with

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the square root sign, then subtracting something else, if

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you want to get rid of that square root sign.

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So what we are going to do is going to multiply-- Essentially

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the conjugate we normally apply to complex numbers.

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But if we have something like a plus b, the conjugate

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is a minus b, right?

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Or if we have something like a minus b, the

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conjugate is a plus b.

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And the reason why-- And what we're going to do is we're

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going to multiply by the conjugate of this.

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And why does-- Why is that normal?

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Why is that useful?

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Because we have a minus b.

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If we multiply it times a plus b, we get a squared

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minus b squared.

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Which will make this radical sign disappear without

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too much work.

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So let's do that.

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Let's multiply by the conjugate of this thing.

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But we can't just multiply it, right?

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We have to multiply it by it over itself.

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Because you can only-- To not change the value of something,

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you can only multiply it by 1.

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So let's multiply it by the conjugate: x squared plus

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4x plus 1 plus x, right?

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That's the conjugate, right?

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Instead of minus x, we have a plus x.

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And we can't just multiply that.

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We have to multiply it by 1, otherwise we would

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be changing the value.

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So it's going to be divided by the same thing.

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x squared plus 4x plus 1 plus x.

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Let me erase this stuff down here so we don't

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get distracted.

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Don't want to get distracted.

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And so what do we have?

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This will become the limit as x approaches infinity.

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Well, this is a minus b times a plus b.

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So we end up with a squared.

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Well, what's this squared?

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That's x squared plus 4x plus 1 minus b squared.

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Well what's b squared?

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Well, b is x.

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So it's going to be minus x squared.

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This is just algebra.

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Divided by this thing: the square root of x squared

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plus 4x plus 1 plus x.

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So let's see.

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There's a little simplification we can do.

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We can subtract-- We can-- These top two terms cancel out.

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So x squared minus x squared.

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And now let's see what we can do.

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Well since we're taking x to infinity, and this is

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what you normally do what you take x to infinity.

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We can divide the numerator and the denominator by

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our highest degree term.

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And in this case, our highest degree terms is x, right?

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We have an x here and an x here.

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And then when you take-- And then, of course, when you

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divide something like this by x, and you take it to,

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infinity, this will approach 0.

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So let's do that.

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Let's divide the numerator and the denominator by x.

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And remember: anything you do in the numerator, you have

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to do in the denominator.

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Otherwise you're changing the value.

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So times 1 over x over 1 over x.

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I'm just dividing the numerator and the denominator by x.

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So this is equal to the limit as x approaches infinity of--

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What's-- That's going to be 4 plus 1 over x over--

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Let me ask you a question.

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What is 1 over x times this thing?

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What is-- This is a bit of an algebra review, but 1 over x.

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What's 1 over x times x squared plus 4x plus 1?

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I'm just doing a little on the side here.

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Well if we take the 1 over x and we put it into the

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radical sign, it becomes 1 over x squared, right?

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This is the same thing as-- You can say 1 squared

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over x squared, but-- 1 over x squared.

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You could say 1 squared.

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You could put a square there, but-- Times x squared

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plus 4x plus 1.

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And that should make sense to you, right?

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Because if we started with this, we could easily just

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take the square root of this and take it outside.

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And the square root of this is 1 over x.

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I'm just going in the other direction, right?

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So, assuming you're comfortable with that, everything

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under the radical sign.

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Even though we're actually dividing by 1 over x, since

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we're going into the radical sign, we're actually

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dividing by x squared.

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So it becomes-- this is the radical sign --x squared

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divided by x squared is 1, right?

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I hope you understand why we're dividing by x squared here.

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We're actually dividing by 1 over x.

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But when you put it under the radical sign, it becomes

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1 over x squared.

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Let me put it this way.

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1 over x times the square root of a.

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That's the same thing.

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That equals the square root of 1 squared over x

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squared times a, right?

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And this is just 1 over x squared.

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So that's the property, or the algebra that we're using.

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So anyway, we divide all of this by x squared.

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So that becomes a 1 plus 4 over x plus 1 over x squared.

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And then, of course, we divide this one by-- This term right

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here, we divide by x, right?

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Because it's not in the radical sign.

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So that just becomes a 1.

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So now what's the limit as x approaches infinity?

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Well as x approaches infinity, this term right here

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goes to 0, right?

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1 over infinity is 0.

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This term right here goes to 0.

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This term right here goes to 0.

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And so what are we left with?

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This is equal to 4 over the square root of 1 plus 1.

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Well that's just 1.

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So that equals 4 over 2, which is equal to 2.

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There you go.

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Now let's do one more problem.

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This is the third one she'd given me.

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In it, we had to kind of brush off our trig identities.

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And really these harder limit problems, they're all about

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kind of knowing your algebra and your trigonometry

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really well.

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Just so you know how to manipulate these functions.

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Because the limit part-- You just have to get it into a form

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where taking the limit is fairly straightforward.

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So let's do that trig problem.

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Clear image.

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Invert colors.

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So it was the limit as x approaches 0

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of cotangent of 2x.

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Was that it?

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Yeah.

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It was cotangent of 2x over the cosecant of x.

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And this one, just like previous problems, more than

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knowing your pre-calculus or your calculus, you need to

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know your trig identities.

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So cosecant of x.

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That's just 1 over sine.

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I remember that by saying it's not intuitive.

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You would have thought that the cosecant is 1 over cosine.

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But it's not.

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It's 1 over sine.

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So I remember that it's not intuitive.

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And cotangent of 2x is equal to 1 over tangent of 2x.

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And tangent is sine over cosine, so cotangent's

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the opposite.

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And so that equals cosine of 2x over sine of 2x, right?

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OK.

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So what is this equal to?

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So this is going to be the limit as x approaches 0.

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Cotangent of 2x, we said, was cosine of 2x over sine of 2x.

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And then it's going to be all of that over the cosecant of x.

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Well that's just 1 over sine of x.

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Well if you divide by 1 over sine of x, that's the same

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thing as multiplying by sine of x.

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So we have-- What do we have?

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We have cosine-- Well we have sine of x times cosine of 2x.

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All of that divided by sine of 2x.

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Just doing a little arithmetic.

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And we have a problem here still.

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Because when you take x approaches 0, this term right

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here goes to 0 and we have a 0 in the denominator, which

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is just not acceptable.

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Because it's undefined.

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And that's the whole reason why we're doing this

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limit to begin with.

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And actually that's the first thing you should have done.

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You should have tried to put it and you would have seen that

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you would've gotten a 0 value in the denominator and it would

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have been unevaluatable.

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285
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Right.

286
00:13:36,169 --> 00:13:37,389
Because really, this is just-- We haven't even

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00:13:37,389 --> 00:13:38,340
done any manipulation yet.

288
00:13:38,340 --> 00:13:40,190
This is the same thing as this.

289
00:13:40,190 --> 00:13:43,870
And if you put the 0 right here, you get undefined.

290
00:13:43,870 --> 00:13:45,039
So what can we do?

291
00:13:45,039 --> 00:13:47,209
Well this is where your break out the trig or you

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00:13:47,210 --> 00:13:48,710
brush off your memory.

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What is the sine of 2x equal?

294
00:13:50,269 --> 00:13:53,490
And this is your double-- One of the double angle formulas.

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00:13:53,490 --> 00:14:01,659
Sine of 2x is equal to 2 sine of x cosine of x.

296
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So if you know that, then you've gone a long way because

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then it becomes pretty simple to simplify.

298
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So it becomes 2 sine of x cosine of x.

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00:14:13,710 --> 00:14:16,710
And if we assume that x isn't 0, it's just approaching 0, we

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00:14:16,710 --> 00:14:21,139
can divide the numerator and the denominator by sine of x.

301
00:14:21,139 --> 00:14:23,340
And what are we left with?

302
00:14:23,340 --> 00:14:29,870
We're left with the limit as x approaches 0 of cosine

303
00:14:29,870 --> 00:14:35,419
of 2x over 2 cosine of x.

304
00:14:35,419 --> 00:14:38,269
Well what's cosine of 0?

305
00:14:38,269 --> 00:14:40,399
Cosine of 0 is 1, right?

306
00:14:40,399 --> 00:14:43,009
So cosine of 2 times 0, which is 0, that's also 1.

307
00:14:43,009 --> 00:14:46,559
So that is equal to 1 over-- right, cosine of 0 is

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00:14:46,559 --> 00:14:48,489
1 --over 2 times 1.

309
00:14:48,490 --> 00:14:50,690
So it equals 1/2.

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There you go.

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I think those are three fairly meaty limit problems.

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And if you know that, you're probably prepared for something

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that your calculus teacher might throw at you.

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See you in a future video.