Learn about Limit Examples w/ brain malfunction on first prob - part 4 Video

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Learn about Limit Examples w/ brain malfunction on first prob - part 4

My cousin Nadia is taking a summer calculus course and she called me last night and she had some limit problem and they were excellent problems 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 memories I hope the answer is work out like they did last night when we’re going over them on the phone. So if I remember correctly the first problem was a limit as z approaches 2 of z² + 2z – 8 all of that over Z4 – 16. So the first thing that you want to do when you try a limit problem is well you just try to substitute the value into, you know maybe there's no problem when z = 2 and you just evaluate the function z = 2. I have to continue this function then the limit as approaches to is going to be the function at two. But you immediately see a problem if you put two right here 24 – 16 and you get the zero in the denominator which is undefined. So we have to figure out someway to get around that. And nine times out of ten when you see a problem like this the solution because both numerator and the denominator look factorable, is to factor that numerator and the denominator. So this is equal to the limit as z approaches 2. What’s the numerator factor? Let see something that when you add two numbers you get positive two and you multiple them you get – 8 so it’s by +4 and –2, right so that’s going to be x + —no not an x I'm so conditioned like a dog. Oh, I can't even undo it anymore, anyway. Well, actually let me erase that, I don’t want to be messy. I tried to undo it and it doesn’t remember everything. So it’s z + 4 × z – 2, that just factoring a quadratic. And what is this? This has the form a² – b², so that’s going to be but a² and if this is a² – b², a would be z² so its z² + 4 × c² – 4 and then of course this also has a form a² + b², right. So this will further factor into z + 2 × z – 2. Well, our factoring paid off, we see a term in the numerator and the denominator that are the same and not only they are the same but if this is the term that was giving us the problem, right because you put it two and then you got zero. So let's assume that we’re not evaluating it at z = 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 arbitrary, as z approaches to of z + 4/z² + 4 × z + 2 which is equal to what? That’s equal to 6 over what's 2² + 4 is 8 and then 2 + 2 is 4, 6/12 is equal to ½. There you go. I thought that was a pretty interesting problem. Let's do another one and this one I've found even more interesting she gave me. It’s really testing my memory to see if I can—but I remember to adjust on the problem so I might get the exact number she’d given me but I forget the exact properties. So it’s the limit as x approaches infinity of the square root and this since square root of x² + 4x +1 – x. So when you look at this you’re like, “Well, let see what's happening here, let’s see this—well as you go to infinity, this term will get really big but the we’re taking the square root of it and it seems like this term will over power this term”. And then you know so maybe, this kind of converted to x but then we would subtract an x so maybe it approaches zero 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 and subtracting something else if you want to get rid of that square root sign. So we are going to do is going to multiply essentially the conjugate we normally apply to complex numbers but you know, if we have something like a + b the conjugate is a – b or if we have something like a – b the conjugate is a + b. And the reason why and what we’re going to do is we're going to multiple by the conjugate of this and why is that normal, why is that useful? Because we have a – b if we multiply it times a + b we get a² – b² which will make this radical sign disappear without too much work. So let's do that let's multiple by the conjugate of this thing. But we can't just multiply it. 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 one. So let's multiply it by conjugate x² + 4x + 1 + x, that’s the conjugate instead of minus x we have a plus x. And we can't just multiple that, we have to multiply it by one otherwise you can changing values so it’s going to be divided by the same thing x² + 4x + 1 + x. Let me erase this stuff down here so we don’t get distracted. I 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 – b × a + b so we end up with a². But what’s this squared? That’s x² + 4x + 1 – b², well what's b²? b is x so that’s going to be -x². This is just algebra—divided by this thing. The square root of x² + 4x + 1 + x. So let see there's a little simplification we can do. We can subtract, we can—these two top two terms cancel out so x² – x². And now let see what we can do. Well, since were taking x to infinity and this is what you normally do when you take x to infinity, we can divide the numerate and the denominator by our highest degree term and in this case our highest degree term is x, right. We have an x here and then an x here and then of course we divide something like this by actually take to infinity, just will approach to zero. So let's do that. Let's divide the numerator and denominator by x. Remember anything you do the numerator you have to do the denominator otherwise you're changing the value, so times one over x over one over x. I'm just dividing the numerator the denominator by x. So this is equal to the limit as the x approaches infinity. That’s going to be 4 + 1/x over—Let me ask you a question, what is 1/x times this thing? What is, and this is a bit of an algebra review but 1/x, what's 1/x × x ² + 4x + 1? I'm just doing a little side here. Well if we take the 1/x and we put it to the radical sign it becomes 1/x². This is the same thing as you can say 1²/x² but 1/x², you could say 1² you can put squared there but times x² + 4x + 1. And that should make sense to you, right because if we started with this we can easily just take the square root of this and take it outside and square root of this 1/x so I'm just going in the other direction. Right? So, assuming your comfortable with that, everything with under the radical sign even though we’re actually dividing by one over x, since we’re going into the radical sign, we’re actually dividing by x² so it becomes, this is the radical sign x² divided by x² is one. I hope you understand why we’re dividing by x² here. We’re actually dividing by 1/x when you put on the radical sign becomes 1/x². Let me put at this way 1/x times the square root of a that’s the same thing that equals the square root of 1²/x² × a and this is just 1/x². So that’s the property or the algebra that we’re using. So anyway, we divide all of these by x². So that becomes a 1+ 4/x + 1/x² and of course we divide this one by this term right here we divide by x because it's not in the radical sign. So that’s just becomes a one. So now what's the limit is x approach infinity? Well as x approaches infinity, this term right here goes to zero, right one over infinity zero. This term right here goes to zero, this term right here goes to zero and so what do we left with? This is equal to 4 over; let’s see the square root of 1 + 1 well that’s just one so that’s equal 4/2 which is equal to two. There you go. Now let's do one more problem. This is the third one she’d given me and we had to kind of brush off our trig identities and really this harder limit problems, they're all about kind of knowing your algebra and your trigonometry really well just so you know how to manipulate this functions. To the limit part you have to get into a form we’re taking a limit is fairly straight forward. So let's do that trig problem. So it was the limit as x approaches zero 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 you trig identities. So cosecant of x that just one over sine, I remember that by saying it's not intuitive you have thought cosecant is one over cosine but it's not, it’s one over sine. So I remember that is not intuitive. And cotangent of 2x is equal to one over tangent of 2x and tangent is sine over cosine, so cotangent is the opposite. So that equals cosine of 2x over sine of 2x. Right? Okay. So what is this equal too? So this is going to be the limit as x approaches zero. 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 just one over sine of x or if you divide by one over sine of x that’s the same thing as multiplying by sine of x. What do we have? We have cosine over sine of x × cosine of 2x, all of that divided by sine of 2x. Just is doing a little arithmetic. And we have a problem here still because when you take x approaches zero this term right here goes to zero and we have the zero then the denominator which is does 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 try to put it and you would have seen that you’ve got a zero value and the denominator and it would have been unevaluatable. Right, because really this just, we haven’t even done and any manipulation yet, this is the same thing as this and if you put the zero right here you get undefined. So what can we do? Well this is where you break out the trig or you brush off your memories. What is the sine of 2x equals? And this is 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 along way because then it becomes pretty simple to simplify. So it becomes 2 sine of x, cosine of x. And then if we assume that x isn’t zero it just approaching zero we can divide to numerator and denominator by sine of x. And what are we left with? We’re left with the limit as x approaches zero of cosine of 2x over 2 cosine of x. What cosine of zero? Cosine of zero is one, right. So cosine of 2 x 0 is zero that’s also one. So that is equal to one over, right cosine of zero is one over two times one. So it equals ½. There you go, I think those are three fairly midi limit problems and if you know that you're probably prepared for something that your calculus teacher might throw at you. See you in the future video.

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My cousin Nadia is taking a summer calculus course and she called me last night and she had some limit problem and they were excellent problems 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 memories I hope the answer is work out like they did last night when we’re going over them on the phone. So if I remember correctly the first problem was a limit as z approaches 2 of z² + 2z – 8 all of that over Z4 – 16. So the first thing that you want to do when you try a limit problem is well you just try to substitute the value into, you know maybe there's no problem when z = 2 and you just evaluate the function z = 2. I have to continue this function then the limit as approaches to is going to be the function at two. But you immediately see a problem if you put two right here 24 – 16 and you get the zero in the denominator which is undefined. So we have to figure out someway to get around that.

And nine times out of ten when you see a problem like this the solution because both numerator and the denominator look factorable, is to factor that numerator and the denominator. So this is equal to the limit as z approaches 2. What’s the numerator factor? Let see something that when you add two numbers you get positive two and you multiple them you get – 8 so it’s by +4 and –2, right so that’s going to be x + —no not an x I'm so conditioned like a dog. Oh, I can't even undo it anymore, anyway. Well, actually let me erase that, I don’t want to be messy. I tried to undo it and it doesn’t remember everything. So it’s z + 4 × z – 2, that just factoring a quadratic. And what is this? This has the form a² – b², so that’s going to be but a² and if this is a² – b², a would be z² so its z² + 4 × c² – 4 and then of course this also has a form a² + b², right. So this will further factor into z + 2 × z – 2.

Well, our factoring paid off, we see a term in the numerator and the denominator that are the same and not only they are the same but if this is the term that was giving us the problem, right because you put it two and then you got zero. So let's assume that we’re not evaluating it at z = 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 arbitrary, as z approaches to of z + 4/z² + 4 × z + 2 which is equal to what? That’s equal to 6 over what's 2² + 4 is 8 and then 2 + 2 is 4, 6/12 is equal to ½. There you go. I thought that was a pretty interesting problem.

Let's do another one and this one I've found even more interesting she gave me. It’s really testing my memory to see if I can—but I remember to adjust on the problem so I might get the exact number she’d given me but I forget the exact properties. So it’s the limit as x approaches infinity of the square root and this since square root of x² + 4x +1 – x. So when you look at this you’re like, “Well, let see what's happening here, let’s see this—well as you go to infinity, this term will get really big but the we’re taking the square root of it and it seems like this term will over power this term”. And then you know so maybe, this kind of converted to x but then we would subtract an x so maybe it approaches zero 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 and subtracting something else if you want to get rid of that square root sign.

So we are going to do is going to multiply essentially the conjugate we normally apply to complex numbers but you know, if we have something like a + b the conjugate is a – b or if we have something like a – b the conjugate is a + b. And the reason why and what we’re going to do is we're going to multiple by the conjugate of this and why is that normal, why is that useful? Because we have a – b if we multiply it times a + b we get a² – b² which will make this radical sign disappear without too much work.

So let's do that let's multiple by the conjugate of this thing. But we can't just multiply it. 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 one. So let's multiply it by conjugate x² + 4x + 1 + x, that’s the conjugate instead of minus x we have a plus x. And we can't just multiple that, we have to multiply it by one otherwise you can changing values so it’s going to be divided by the same thing x² + 4x + 1 + x.

Let me erase this stuff down here so we don’t get distracted. I 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 – b × a + b so we end up with a². But what’s this squared? That’s x² + 4x + 1 – b², well what's b²? b is x so that’s going to be -x². This is just algebra—divided by this thing. The square root of x² + 4x + 1 + x. So let see there's a little simplification we can do. We can subtract, we can—these two top two terms cancel out so x² – x².

And now let see what we can do. Well, since were taking x to infinity and this is what you normally do when you take x to infinity, we can divide the numerate and the denominator by our highest degree term and in this case our highest degree term is x, right. We have an x here and then an x here and then of course we divide something like this by actually take to infinity, just will approach to zero. So let's do that.

Let's divide the numerator and denominator by x. Remember anything you do the numerator you have to do the denominator otherwise you're changing the value, so times one over x over one over x. I'm just dividing the numerator the denominator by x. So this is equal to the limit as the x approaches infinity. That’s going to be 4 + 1/x over—Let me ask you a question, what is 1/x times this thing? What is, and this is a bit of an algebra review but 1/x, what's 1/x × x ² + 4x + 1? I'm just doing a little side here.

Well if we take the 1/x and we put it to the radical sign it becomes 1/x². This is the same thing as you can say 1²/x² but 1/x², you could say 1² you can put squared there but times x² + 4x + 1. And that should make sense to you, right because if we started with this we can easily just take the square root of this and take it outside and square root of this 1/x so I'm just going in the other direction. Right?

So, assuming your comfortable with that, everything with under the radical sign even though we’re actually dividing by one over x, since we’re going into the radical sign, we’re actually dividing by x² so it becomes, this is the radical sign x² divided by x² is one. I hope you understand why we’re dividing by x² here. We’re actually dividing by 1/x when you put on the radical sign becomes 1/x². Let me put at this way 1/x times the square root of a that’s the same thing that equals the square root of 1²/x² × a and this is just 1/x². So that’s the property or the algebra that we’re using.

So anyway, we divide all of these by x². So that becomes a 1+ 4/x + 1/x² and of course we divide this one by this term right here we divide by x because it's not in the radical sign. So that’s just becomes a one. So now what's the limit is x approach infinity? Well as x approaches infinity, this term right here goes to zero, right one over infinity zero. This term right here goes to zero, this term right here goes to zero and so what do we left with? This is equal to 4 over; let’s see the square root of 1 + 1 well that’s just one so that’s equal 4/2 which is equal to two. There you go.

Now let's do one more problem. This is the third one she’d given me and we had to kind of brush off our trig identities and really this harder limit problems, they're all about kind of knowing your algebra and your trigonometry really well just so you know how to manipulate this functions. To the limit part you have to get into a form we’re taking a limit is fairly straight forward. So let's do that trig problem.

So it was the limit as x approaches zero 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 you trig identities. So cosecant of x that just one over sine, I remember that by saying it's not intuitive you have thought cosecant is one over cosine but it's not, it’s one over sine. So I remember that is not intuitive. And cotangent of 2x is equal to one over tangent of 2x and tangent is sine over cosine, so cotangent is the opposite. So that equals cosine of 2x over sine of 2x.

Right? Okay. So what is this equal too? So this is going to be the limit as x approaches zero. 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 just one over sine of x or if you divide by one over sine of x that’s the same thing as multiplying by sine of x. What do we have? We have cosine over sine of x × cosine of 2x, all of that divided by sine of 2x. Just is doing a little arithmetic.

And we have a problem here still because when you take x approaches zero this term right here goes to zero and we have the zero then the denominator which is does 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 try to put it and you would have seen that you’ve got a zero value and the denominator and it would have been unevaluatable. Right, because really this just, we haven’t even done and any manipulation yet, this is the same thing as this and if you put the zero right here you get undefined.

So what can we do? Well this is where you break out the trig or you brush off your memories. What is the sine of 2x equals? And this is 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 along way because then it becomes pretty simple to simplify. So it becomes 2 sine of x, cosine of x. And then if we assume that x isn’t zero it just approaching zero we can divide to numerator and denominator by sine of x.

And what are we left with? We’re left with the limit as x approaches zero of cosine of 2x over 2 cosine of x. What cosine of zero? Cosine of zero is one, right. So cosine of 2 x 0 is zero that’s also one. So that is equal to one over, right cosine of zero is one over two times one. So it equals ½. There you go, I think those are three fairly midi limit problems and if you know that you're probably prepared for something that your calculus teacher might throw at you. See you in the future video.

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