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Calculus: L'Hopital's Rule, Indeterminate Products


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About this Lesson

  • Type: Video Tutorial
  • Length: 7:47
  • Media: Video/mp4
  • Use: Watch Online & Download
  • Access Period: Unrestricted
  • Download: MP4 (iPod compatible)
  • Size: 83 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)

Some indeterminate forms, including indeterminate products, must be manipulated before L'Hôpital's Rule can be applied. In order to do this, try expressing one of the factors as a fraction (so you can recognize things like (0 * infinity) or (theta * cotangent of theta) as an indeterminate form given that they are actually indeterminate products rather than the easily recognizable 0/0 or infinity/infinity). Once recognized as an indeterminate form, you will learn how to manipulate these camouflaged indeterminate forms in such a way that they can have L'Hôpital's rule applied to them. In this lesson, you will learn how to recognize these limits with indeterminate forms as well as how to handle limits with negative exponents and limits with trigonometric expressions.

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 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.

About this Author

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L’Hospital’s Rule
Other Indeterminate Forms
L’Hospital’s Rule and Indeterminate Products Page [1 of 2]
Okay, now I really should be in camouflaged fatigues for this one, because what I want to tell you about are things that I consider to be camouflaged indeterminate forms. So what does that mean? Well, remember that indeterminate form is some limit that, when you sort of plug in the limit, you're seeing either ?/? or 0/0, that kind of thing. So how can I camouflage that in some sort of exotic way? Well, let’s take a look at some examples and see that, in fact, even things like zero over zero can be camouflaged.
So how about the limit as x approaches infinity of e-x ln x? So there’s a lot of complicated functions there. We have the exponential function there and the logarithm function, all there in one fell swoop. So again, we take a look at this and then we think, “What’s going on here?” Well, if x is approaching infinity, certainly this thing, I think it’s okay to see the natural log goes to infinity. So what I'm seeing here is infinity. So that’s okay. And what about e-x? Well, let’s think about this for a second. If x is getting really, really big, then I have e- and a really, really big number. So what is e-7, e-100, e-1000? Well, the negative sign puts me underneath, it’s a flip, and so I'm getting smaller and smaller and smaller. So, in fact, this limit right here of this piece is actually 0. So, in fact, I see something that looks like this: 0 × ?. And that actually is an indeterminate form, even though it’s not the garden-variety 0/0 or ?/?. This is sort of weird. It’s 0 × ?, and you can sort of see why it should be an indeterminate form. Who wins out? 0 times anything - that should be 0. On the other hand, anything times ? should be ?. So you can see it’s indeterminate. Who knows what the answer is?
Well, one way to get around this is to uncamouflage this, expose it for what it really is. Now, what I'm about to show you here is absolutely fantasy math. If you tell anyone else that I said this, I will deny it. You understand? This is just between us. But look, let’s think about this. What is zero? Zero, if you cut me some slack, is really nothing more than 1/?. If something is going 1/?, that’s sort of going to 0. So, in fact, 0 is just 1/?. If I think of it that way, then really I've got the old-fashioned indeterminate form infinity over infinity. So the trick is to realize that if you see something that’s approaching 0 × ?, I can take that zero and write it as a 1 over something that’s going to ?. So that is a great technique to uncover this camouflage limit. Let’s apply that idea here.
So instead of writing e-x, I'm going to write it as 1 over something. Now, what would it be? Well, it would be . That negative sign means I write it as a 1 over. So this would be . Think about that. What does this equal? Well, since I have a 1 over, that means I have a negative exponent, and so there it is, negative x. So these two things are actually equal. And then I have the ln x, so I just took care of that negative sign by writing it as . And now, I can write this limit as . Great! Because now notice that as x goes to infinity, I see, in fact, infinity on top and I see infinity on the bottom. So this is now an indeterminate form of the garden variety ?/?, so I can apply L’Hôspital’s Rule. I can’t apply L’Hôspital’s Rule right here. I've got to bring it into the form that I know. That’s why it’s camouflaged. So what does this equal? It equals the limit, as x approaches infinity – if I take the derivative of the top, and that gives me , and the derivative of the bottom, the derivative of ex, that’s great, because it’s just itself, ex. So what does that equal? That equals the limit as x approaches infinity of – and that , I could write that x down here, the 1 over, way down here, so it migrates down – x ex. And what’s that limit? Well, as x goes to infinity, this thing is blowing up, so 1 over something going to infinity is approaching zero. So, in fact, this limit does exist and equals zero. We didn’t know that first looking at it, because we had a funny camouflaged indeterminate form.
All right, let’s try one last one. Again, in the genre of camouflaged linear forms, let’s take the limit as theta approaches zero of theta cotangent of theta. So I'm using thetas now instead of x’s. It doesn’t bother us. Who cares, it’s just a variable. And I've got the cotangent here. So that’s a little exotic. Let’s see what goes on. Well, as ? approaches 0, well, theta’s approaching zero. So that’s just a zero-thing, so there’s 0. And what about cotangent? Well, as ? approaches 0, cotangent actually is approaching an asymptote. So, in fact, that thing is actually shooting off to infinity. So, in fact, this piece is going to infinity. So, once again, I see a camouflaged indeterminate form. It’s an indeterminate form, but not of the flavor that we prefer to think about. So what would I do here? Well, one technique here is to take the infinity and write it as 1 over zero. See, infinity is really 1/0. Not really, don’t tell anyone, this is just between us. But think about it. 1/0 is sort of heading toward infinity. So if I could somehow rewrite this infinity as 1/0, then I really do have my 0/0 classic indeterminate form and I can apply L’Hôspital‘s Rule. So let’s see how I can do that.
So without changing anything at all, no slight of hand here, I want to write this as a 1 over something. Well, what is cotangent? Well, cotangent is . So, in fact, without changing anything at all, I can just rewrite this as , because, in fact, that’s the definition of cotangent, . And so this equals the . And now, what’s that limit? Well, as ? goes to 0, I get 0 on the top and tan 0 = 0, and so I'm left with an indeterminate form of the garden variety, 0/0. Great, now I can apply L’Hôspital’s Rule. So if I apply L’Hôspital’s Rule, I see the limit as ? approaches 0 and the derivative of ? = 1, the derivative of tan ? = sec2?, and again, if you like secants, then you're in good shape. All I know about secant is that it’s 1/cos, so 1/sec is just cos2?. And if I take the limit as ? approaches 0, then what do I see? Well, cos ? is approaching, and so 12 is 1. So this limit, in fact, does exist and equals 1.
So you can see that if you see something that looks like zero times infinity, it really is a camouflage 0/0 or ?/?. Just figure out a way of decoding that, and then once you’ve got the standards 0/0, ?/?, you can use L’Hôspital’s Rule. See ya.

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