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Calculus:L'Hopital's Rule-Indeterminate Difference

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About this Lesson

  • Type: Video Tutorial
  • Length: 14:38
  • Media: Video/mp4
  • Use: Watch Online & Download
  • Access Period: Unrestricted
  • Download: MP4 (iPod compatible)
  • Size: 158 MB
  • Posted: 11/18/2008

This lesson is part of the following series:

Calculus (279 lessons, $198.00)
Calculus Review (48 lessons, $95.04)
Calculus: L'Hopital's Rule (8 lessons, $11.88)
Calculus: Other Indeterminate Forms (4 lessons, $6.93)

This lesson will show you how to recognize indeterminate forms that are camouflaged as indeterminate differences rather than showing up as 0/0 or infinity/infinity. For instance, when you calculate the limit to be equal to infinity - infinity (e.g. 1/x - 1/(e^x-1)), this is an indeterminate form, so you must manipulate it to make it into a form such that you can apply L'Hôpital's rule to it. This lesson also covers additional camouflaged indeterminate differences that show up with fractions and radicals. To apply L'Hôpital's rule to an indeterminate difference in many of these instances, you will first have to find a common denominator or look for a sneaky way to factor. Professor Burger will walk you through examples of all of these instances.

Taught by Professor Edward Burger, this lesson was selected from a broader, comprehensive course, College Algebra. This course and others are available from Thinkwell, Inc. The full course can be found at http://www.thinkwell.com/student/product/calculus. The full course covers limits, derivatives, implicit differentiation, integration or antidifferentiation, L'Hôpital's Rule, functions and their inverses, improper integrals, integral calculus, differential calculus, sequences, series, differential equations, parametric equations, polar coordinates, vector calculus and a variety of other AP Calculus, College Calculus and Calculus II topics.

Edward Burger, Professor of Mathematics at Williams College, earned his Ph.D. at the University of Texas at Austin, having graduated summa cum laude with distinction in mathematics from Connecticut College.

He has also taught at UT-Austin and the University of Colorado at Boulder, and he served as a fellow at the University of Waterloo in Canada and at Macquarie University in Australia. Prof. Burger has won many awards, including the 2001 Haimo Award for Distinguished Teaching of Mathematics, the 2004 Chauvenet Prize, and the 2006 Lester R. Ford Award, all from the Mathematical Association of America. In 2006, Reader's Digest named him in the "100 Best of America".

Prof. Burger is the author of over 50 articles, videos, and books, including the trade book, Coincidences, Chaos, and All That Math Jazz: Making Light of Weighty Ideas and of the textbook The Heart of Mathematics: An Invitation to Effective Thinking. He also speaks frequently to professional and public audiences, referees professional journals, and publishes articles in leading math journals, including The Journal of Number Theory and American Mathematical Monthly. His areas of specialty include number theory, Diophantine approximation, p-adic analysis, the geometry of numbers, and the theory of continued fractions.

Prof. Burger's unique sense of humor and his teaching expertise combine to make him the ideal presenter of Thinkwell's entertaining and informative video lectures.

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L’Hospital’s Rule
Other Indeterminate Forms
L’Hospital’s Rule and Indeterminate Differences Page [1 of 3]
All right, let’s take a look at some even more camouflaged indeterminate forms and see if we can make head or tail out of these things.
So, first of all, let’s just consider the . Now, that looks really peculiar, because this is one of the instances where you don’t even have a multiplication or division going on here. We also have subtraction of these guys. So let’s see what happens if we actually try to take the limit first, always the first step in any good limit question.
So, if x ? 0, ? ?. So we have infinity here, and then minus, and then what happens here? If x approaches 0, then I have 1 over e to the – and that’s approaching 0, so that’s e0. e0 = 1, and so I have 1 –1 = 0. So I see 1/0, which is another infinity. So what I see is a very unusual indeterminate form. This is an exotic indeterminate form, an indeterminate form of the garden-variety infinity minus infinity. What is that? Well, maybe this infinity is so big that even subtracting that infinity keeps it infinity. Maybe this infinity is so big that this infinity minus that infinity makes negative infinity. Maybe they're the same exact frequency of infinity, so everything cancels out and you get 0. Who knows? It’s an indeterminate form. This needs more work. Now, we just can’t apply L’Hôspital’s Rule, because L’Hôspital’s Rule is a fact that says if you’ve got something divided by something, then you take the derivative of the top and the derivative of the bottom. It’s all this complicated stuff. So, what to do?
Well, it’s again the technique of massaging this into a form that I can apply L’Hôspital’s Rule. So let’s see how I can massage. Well, massages are always great, they’re very relaxing. Now here the obvious massage is just to make this into one big fraction. This is crying out to be combined into one fraction. Who would write something like this? You’d want one fraction, so let’s get a common denominator and combine these. So here I have to multiply top and bottom by ex-1, and then here, top and bottom, I have to multiply by x. So if I do that – in fact, let me just do this on the side here so we can calculate this just on our own. So I’d have . You’ll notice that all I did there was get a common denominator. So these terms here, that and that, cancel. I’m just left with 1 over x minus – and then these terms cancel, I’m just left with . Same thing, but now I’ve got the same bottom, so I can combine the tops. If I combine the tops, I see ex - 1 - x, all over that common bottom. So I’ll write that in right here, since that really is what this thing equals after you’ve combined all that. And I see . Okay, that calculation is done. And now, let’s take the limit and see what we get. Notice it is one fraction, so if we only got 0/0 or ?/?, we could use L’Hôspital’s Rule. So let’s hope.
Okay, back to approaching zero, this thing here, e0, that’s 1. So 1 - 1 = 0, and then -0, 0 on top. We’re halfway home. What about the bottom? Well, that’s 0 times who cares what, 0 times anything over here = 0, so I see 0/0. This is, in fact, an indeterminate form of our favorite flavor, 0/0, so I can now use L’Hôspital’s Rule. Great! See how I converted the ? - ? to something like this, 0/0? That’s an idea and that’s a technique that will work. So let’s see what happens if we now apply L’Hôspital’s Rule. Take the derivative of the top, so I just go by and do it by pieces, so I see that the derivative of ex is ex, the derivative of –1 = 0, the derivative of -x = -1. Now, the bottom is going to require a product rule. Now, you may say, “Hey, wait a minute. What if I distribute first and write this as x ex - x?” You can do that, but it’s still going to require a product rule right in there. So I’ll just use the product rule right off the bat. So that’s the first times the derivative of the second, and the derivative of this is just ex, plus the second ex - 1, multiplied by the derivative of the first, which is just 1. So, in fact, that’s the derivative of the bottom. It looks a little bit complicated, I admit, but let’s see what happens if we take the limit.
If we take the limit, what do I see? Well, I see as x goes to zero, this goes to e0, so that’s 1 - 1, that’s 0 on top. And here, this is going to0 , because this whole term here, that’s 0 , so 0 times anything is 0 , so that’s 0. And then I have e0, which is 1, so 1 - 1, so that’s 0. So I have 0/0 again. Limit doesn’t exist? No, it’s another indeterminate form. Are we making any progress? I don’t know. But the thing to do is to try to use L’Hôspital’s Rule another time. You may be wondering, in fact, you can sort of, if you like to wager, you might want to wager how many times could one use L’Hôspital’s Rule? Well, a lot. So let’s take a look and try it using L’Hôspital’s Rule again.
The derivative of the top is just ex. Now, the bottom is going to be complicated. First of all, I've got a little product rule right here, so that’s the first x times the derivative of ex, which happily is itself, plus the second, which is ex, times the derivative of x, which is happily just 1, plus the derivative of ex, that’s just itself, and then the derivative of -1 = 0. Okay, well, let’s take this limit and see what happens.
Well, on the top, as x approaches 0, notice this thing now does not approach 0, it approaches e0, which is 1. So no matter what happens, we know we’re not going to have an indeterminate form, because on the top I’m approaching 1. Now what about on the bottom? Well, this is approaching 0, because I have that term there, which is 0, so that’s 0. This is e0 plus e0. Well, that’s 1 + 1, which is 2. So, in fact, this limit does exist and equals 2. Doesn’t equal 2, it equals the reciprocal of 2, also known as ½. So there’s a more elaborate example, where, in fact, you can actually see that the difference, after it worked at, the limit can be taken. So that’s great.
Okay, how about another one even more exotic. So you’re saying, “How can it get more exotic, Ed? Can you really make this more exotic?” And the answer is I sure can. Let’s take the limit, as x approaches infinity, of the + x – but I’m not done yet – -2x. Look at that. So that’s got square roots, and it’s got stuff and who knows? Let’s take the limit and see what happens.
Well, as x approaches infinity, this term here is getting really, really big. And the square root of really, really big is still really, really big. This is infinity and this here is also really big, so I’ve got infinity minus infinity, I’ve got this camouflaged person. So what do I do? Well, in this case, what I'm going to do is look at this and ponder. Okay, well, that didn’t work, so what else could I do? Well, here’s sort of a neat trick to sort of lasso this problem. One neat trick is to say, “Well, let me try to factor out this x2 term here.” So let me try to factor out that x2 term. Now, why would I try to factor out that x2 term? Well, because that’s sort of the highest degree here and, if I can pull that out – let’s just think about that for a second. If I could pull out a factor of x2, then if I took the square root of it, the would be an x, and then I would have an x and an x and I would be able to factor this. This is a really sneaky trick, and the trick is see if you can factor out the highest power of x here. Now, let’s see what happens if we try this. And this is generally a trick by the way. This is something that I did not think of, but turns out is a clever trick that often works. So, if you didn’t think of it, then you're in the same boat with me. I'm going to factor out an x2. Now this is going to be tricky. If I factor it out of here, then I’m just left with a 4. But if I factor it out of here, what am I left with? Well, here I’m left with – well, when I multiply x2 by whatever it is, I have to get x. And so that answer has to be 1/x. And you can check that distribute and you’ll see here 4x2 and you’ll see here an x. So that’s great, that’s great. And then, I’ll do nothing here. And it still looks pretty hopeless, I admit. It looks pretty hopeless. So what I’ll do now is notice that I can take the square root of this term and the square root of that term of that term separately. And the square root of x2 is just x, since x is positive. x is going off to infinity, so it’s positive. Can you tell that my black pen is slowly dying, well death is part of life, so it’s okay. But maybe it’ll be resurrected with a new black pen, but we’ll see. Okay, so here we see an x and here we see a . Great! Now, what could I do? Well now I could factor out these x’s. Still not clear why I’d want to do this, but one final step and we’ll see why this trick is really pretty neat. And the black pen is dying. It’s a slow, painful death for both the black pen and me, apparently. Now, let’s take a look at that.
Well, this term here is now going to infinity, as x approaches infinity. And what about this term here? Well, as x approaches infinity, I have 1 over something that’s going to infinity, so that’s 0. That’s approaching zero, this is getting really, really small, and so what I see here is the + 0. So that’s the , which is 2. 2 - 2 = 0. So I’ve got 0 × ?. Hey! Finally, something that’s recognizable. It’s the zero times infinity thing. So, by factoring out that highest power, I was able to rewrite this as a zero times infinity thing. And how do we deal with that? Well, I can just invert this and try to write this out in a different way. Let’s pick the action up right here. In fact, you know what I’ll do? I’ll take this here and I’ll pop it right over there. There it is. Now, if that’s out of the way, what I see is the following. That will equal the limit as x approaches infinity of – now, I've got the top thing still there. I want that infinity thing still there, so that’s going to be . But now I want to take that x in front and write it as a 1 over. But I can’t change the value of it, so I’ve got to write it as a 1 over 1 over. So it comes like that. Now, let’s just pause there and make sure everyone sees that. That’s really just multiplying by x. Watch, if I invert and multiply, the x comes out in front. So this is the same thing as what’s over there, it’s the same thing as this x times this, but now I’m writing it this way. And look what happens when you take the limit. When you take the limit here, what do I see? Well, on the top I'm approaching - 2, which is 0, and on the bottom I'm approaching 0. So now, finally, I'm approaching my favorite 0/0. So now I can finally apply L’Hôspital’s Rule. So let’s do that and see what we get.
x approaches infinity. I’ve got to take the derivative of that. Well, that actually is going to require a chain rule. I've got an inside, I've got junk here, and then I've got an outside square root. So what’s the derivative of ? Well, you could think of the , and so that’s going to . So that’s 4 + , and then the derivative of -2 is just 0. And then we divide that – oh, wait, hold the presses! If you’re going to do the chain rule, we have to be really, really careful. Were you careful? Did you catch what I just did? I made, actually, a mistake. Let’s take a look at this and make sure that we see it. The chain rule requires us to look at the inside and outside. Now look, if I've got , the derivative of that is . But now I’ve got to multiply that by the derivative of the inside, by the derivative of the blop. And the derivative of 4 = 0 and the derivative of 1/x, that’s just x-1, is –1/x2. So now, that’s the correct derivative of this piece. And the derivative of -2 remains 0, so that’s fine. And what’s the derivative of this piece down here? The derivative of that piece down there is –1/x2. And so, happily, I get some cancellation. This whole term cancels with that whole term. And what happens as x races off to infinity? As x races off to infinity, this term right here, this 1/x term, races off to 0. And so, what am I left with? Well, if we just wait for the dust to settle, I'm left with 1/2 + 0, which is just the , and so that’s just 2. So 2 × 2 = 4. So after all that hard work and toil, we see that this original limit, which was just infinity minus infinity upon first glance, after a lot of careful gymnastics, it turns out that limit is approaching ¼. Pretty cool, but the bottom line is look for these little sneaky techniques to convert a sort of unusual indeterminate form into your garden variety favorite indeterminate form. I’ll see you at the next lecture.

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