Limit basics
Limit Examples w/ brain malfunction on first prob (part 4) 3 interesting limit examples (correct answer for problem 1 is 3/16 (6/(4*8) NOT 6/(4+8))
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- My cousin Nadia is taking a summer calculus course and
- she called me last night.
- And she had some limit problems and they were
- excellent problem.
- So I thought it was worthwhile to make videos on the problems
- that she had to figure out.
- Anyway, so let's do them.
- If I remember them correctly.
- I'm doing this on memory.
- So I hope the answers work out like they did last night when
- we were going over them on the phone.
- So if I remember correctly, the first one was the limit
- as z approaches 2 of x.
- Oh no.
- Not that-- There's no x.
- It's a z.
- Of z squared plus 2z minus 8.
- All of that over z to the fourth minus 16.
- So the first thing when you want to do when you try a limit
- problem is, well you just try to substitute the value into--
- Maybe there's no problem when z equals 2 and you just evaluate
- the function at z equals 2.
- If it's a continuous function, then the limit as z
- approaches 2 is going to be the function at 2.
- But you immediately see a problem if you put 2 right
- here-- 2 to the fourth minus 16 --and you get a 0 in the
- denominator, which is undefined, so we have to
- figure out some way to get around that.
- And nine times out of 10, when you see a problem like this,
- the solution, because both the numerator and the denominator
- look factorable, is to factor the numerator and
- the denominator.
- So this is equal to the limit as z approaches 2.
- What's the numerator factored?
- Let's see.
- Something when you add two numbers, you get positive
- 2, and you multiply them, you get minus 8.
- So it's probably plus 4 and minus 2, right?
- So this is going to be x plus-- No.
- Not an x.
- I'm so conditioned.
- I'm like a dog.
- Oh.
- I can't even undo it anymore.
- Anyway, well z-- Actually let me erase that.
- I don't want to be messy.
- Let me erase.
- I tried to undo it and it doesn't remember everything.
- So it's z plus 4 times z minus 2.
- That's just factoring a quadratic.
- And what is this?
- This has the form a squared minus b squared, right?
- So that's going to be-- But a squared-- If this is a squared
- minus b squared, a would be z squared, right?
- So it's z squared plus 4 times z squared minus 4.
- And then, of course, this also has the form a squared
- plus b squared, right?
- So this will further factor into z plus 2 times z minus 2.
- Well, our factoring paid off.
- We see a term in the numerator and the denominator
- that are the same.
- And not only are they the same, but they-- This is
- the term that was giving us the problems, right?
- Because you put a 2 in here, you got a 0.
- So let's assume that we're not evaluating it at z equals 2.
- And so, for all other values, we can divide those
- because those are going to be the same values.
- And then what are we left with?
- This is equal to the limit-- and I've change colors
- arbitrarily --as z approaches 2 of z plus 4 over z squared
- plus 4 times z plus 2.
- Which is equal what?
- That's equal to 6 over-- what's 2 squared plus 4?
- Is 8.
- And then 2 plus 2 is 4.
- 6 over 12 is equal to 1/2.
- There you go.
- I thought that was a pretty interesting problem.
- Let's do another one.
- And this one I found even more interesting.
- She gave me-- It's really testing my memory to see
- if I can-- But I remember the gist of the problem.
- So I might not get the exact numbers she'd given me, but
- hopefully I get the exact properties.
- So it's the limit as x approaches infinity of the
- square root of x squared plus 4x plus 1 minus x.
- So when you look at this you're like well let's see.
- What's happening here?
- Let's see this.
- As you go to infinity, this term will get really big.
- But then we're taking the square root of it.
- And it seems like this term would overpower this term.
- And then, so maybe this kind of converges to x, but then
- we would subtract an x.
- So maybe it approaches 0.
- So that's, at least, that was my first intuition
- when I spoke to her.
- But as we will see, the intuition here is wrong.
- And really to do this you have to know a little
- bit of a trick.
- And this trick pays off a lot whenever you see something with
- the square root sign, then subtracting something else, if
- you want to get rid of that square root sign.
- So what we are going to do is going to multiply-- Essentially
- the conjugate we normally apply to complex numbers.
- But if we have something like a plus b, the conjugate
- is a minus b, right?
- Or if we have something like a minus b, the
- conjugate is a plus b.
- And the reason why-- And what we're going to do is we're
- going to multiply by the conjugate of this.
- And why does-- Why is that normal?
- Why is that useful?
- Because we have a minus b.
- If we multiply it times a plus b, we get a squared
- minus b squared.
- Which will make this radical sign disappear without
- too much work.
- So let's do that.
- Let's multiply by the conjugate of this thing.
- But we can't just multiply it, right?
- We have to multiply it by it over itself.
- Because you can only-- To not change the value of something,
- you can only multiply it by 1.
- So let's multiply it by the conjugate: x squared plus
- 4x plus 1 plus x, right?
- That's the conjugate, right?
- Instead of minus x, we have a plus x.
- And we can't just multiply that.
- We have to multiply it by 1, otherwise we would
- be changing the value.
- So it's going to be divided by the same thing.
- x squared plus 4x plus 1 plus x.
- Let me erase this stuff down here so we don't
- get distracted.
- Don't want to get distracted.
- And so what do we have?
- This will become the limit as x approaches infinity.
- Well, this is a minus b times a plus b.
- So we end up with a squared.
- Well, what's this squared?
- That's x squared plus 4x plus 1 minus b squared.
- Well what's b squared?
- Well, b is x.
- So it's going to be minus x squared.
- This is just algebra.
- Divided by this thing: the square root of x squared
- plus 4x plus 1 plus x.
- So let's see.
- There's a little simplification we can do.
- We can subtract-- We can-- These top two terms cancel out.
- So x squared minus x squared.
- And now let's see what we can do.
- Well since we're taking x to infinity, and this is
- what you normally do what you take x to infinity.
- We can divide the numerator and the denominator by
- our highest degree term.
- And in this case, our highest degree terms is x, right?
- We have an x here and an x here.
- And then when you take-- And then, of course, when you
- divide something like this by x, and you take it to,
- infinity, this will approach 0.
- So let's do that.
- Let's divide the numerator and the denominator by x.
- And remember: anything you do in the numerator, you have
- to do in the denominator.
- Otherwise you're changing the value.
- So times 1 over x over 1 over x.
- I'm just dividing the numerator and the denominator by x.
- So this is equal to the limit as x approaches infinity of--
- What's-- That's going to be 4 plus 1 over x over--
- Let me ask you a question.
- What is 1 over x times this thing?
- What is-- This is a bit of an algebra review, but 1 over x.
- What's 1 over x times x squared plus 4x plus 1?
- I'm just doing a little on the side here.
- Well if we take the 1 over x and we put it into the
- radical sign, it becomes 1 over x squared, right?
- This is the same thing as-- You can say 1 squared
- over x squared, but-- 1 over x squared.
- You could say 1 squared.
- You could put a square there, but-- Times x squared
- plus 4x plus 1.
- And that should make sense to you, right?
- Because if we started with this, we could easily just
- take the square root of this and take it outside.
- And the square root of this is 1 over x.
- I'm just going in the other direction, right?
- So, assuming you're comfortable with that, everything
- under the radical sign.
- Even though we're actually dividing by 1 over x, since
- we're going into the radical sign, we're actually
- dividing by x squared.
- So it becomes-- this is the radical sign --x squared
- divided by x squared is 1, right?
- I hope you understand why we're dividing by x squared here.
- We're actually dividing by 1 over x.
- But when you put it under the radical sign, it becomes
- 1 over x squared.
- Let me put it this way.
- 1 over x times the square root of a.
- That's the same thing.
- That equals the square root of 1 squared over x
- squared times a, right?
- And this is just 1 over x squared.
- So that's the property, or the algebra that we're using.
- So anyway, we divide all of this by x squared.
- So that becomes a 1 plus 4 over x plus 1 over x squared.
- And then, of course, we divide this one by-- This term right
- here, we divide by x, right?
- Because it's not in the radical sign.
- So that just becomes a 1.
- So now what's the limit as x approaches infinity?
- Well as x approaches infinity, this term right here
- goes to 0, right?
- 1 over infinity is 0.
- This term right here goes to 0.
- This term right here goes to 0.
- And so what are we left with?
- This is equal to 4 over the square root of 1 plus 1.
- Well that's just 1.
- So that equals 4 over 2, which is equal to 2.
- There you go.
- Now let's do one more problem.
- This is the third one she'd given me.
- In it, we had to kind of brush off our trig identities.
- And really these harder limit problems, they're all about
- kind of knowing your algebra and your trigonometry
- really well.
- Just so you know how to manipulate these functions.
- Because the limit part-- You just have to get it into a form
- where taking the limit is fairly straightforward.
- So let's do that trig problem.
- Clear image.
- Invert colors.
- So it was the limit as x approaches 0
- of cotangent of 2x.
- Was that it?
- Yeah.
- It was cotangent of 2x over the cosecant of x.
- And this one, just like previous problems, more than
- knowing your pre-calculus or your calculus, you need to
- know your trig identities.
- So cosecant of x.
- That's just 1 over sine.
- I remember that by saying it's not intuitive.
- You would have thought that the cosecant is 1 over cosine.
- But it's not.
- It's 1 over sine.
- So I remember that it's not intuitive.
- And cotangent of 2x is equal to 1 over tangent of 2x.
- And tangent is sine over cosine, so cotangent's
- the opposite.
- And so that equals cosine of 2x over sine of 2x, right?
- OK.
- So what is this equal to?
- So this is going to be the limit as x approaches 0.
- Cotangent of 2x, we said, was cosine of 2x over sine of 2x.
- And then it's going to be all of that over the cosecant of x.
- Well that's just 1 over sine of x.
- Well if you divide by 1 over sine of x, that's the same
- thing as multiplying by sine of x.
- So we have-- What do we have?
- We have cosine-- Well we have sine of x times cosine of 2x.
- All of that divided by sine of 2x.
- Just doing a little arithmetic.
- And we have a problem here still.
- Because when you take x approaches 0, this term right
- here goes to 0 and we have a 0 in the denominator, which
- is just not acceptable.
- Because it's undefined.
- And that's the whole reason why we're doing this
- limit to begin with.
- And actually that's the first thing you should have done.
- You should have tried to put it and you would have seen that
- you would've gotten a 0 value in the denominator and it would
- have been unevaluatable.
- Right.
- Because really, this is just-- We haven't even
- done any manipulation yet.
- This is the same thing as this.
- And if you put the 0 right here, you get undefined.
- So what can we do?
- Well this is where your break out the trig or you
- brush off your memory.
- What is the sine of 2x equal?
- And this is your double-- One of the double angle formulas.
- Sine of 2x is equal to 2 sine of x cosine of x.
- So if you know that, then you've gone a long way because
- then it becomes pretty simple to simplify.
- So it becomes 2 sine of x cosine of x.
- And if we assume that x isn't 0, it's just approaching 0, we
- can divide the numerator and the denominator by sine of x.
- And what are we left with?
- We're left with the limit as x approaches 0 of cosine
- of 2x over 2 cosine of x.
- Well what's cosine of 0?
- Cosine of 0 is 1, right?
- So cosine of 2 times 0, which is 0, that's also 1.
- So that is equal to 1 over-- right, cosine of 0 is
- 1 --over 2 times 1.
- So it equals 1/2.
- There you go.
- I think those are three fairly meaty limit problems.
- And if you know that, you're probably prepared for something
- that your calculus teacher might throw at you.
- See you in a future video.
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At 5:31, how is the moon large enough to block the sun? Isn't the sun way larger?
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