Hi! We show you're using Internet Explorer 6. Unfortunately, IE6 is an older browser and everything at MindBites may not work for you. We recommend upgrading (for free) to the latest version of Internet Explorer from Microsoft or Firefox from Mozilla.
Click here to read more about IE6 and why it makes sense to upgrade.

Calculus: Example of One to the Infinite Power


Like what you see? Buy now to watch it online or download.

You Might Also Like

About this Lesson

  • Type: Video Tutorial
  • Length: 7:53
  • Media: Video/mp4
  • Use: Watch Online & Download
  • Access Period: Unrestricted
  • Download: MP4 (iPod compatible)
  • Size: 84 MB
  • Posted: 06/26/2009

This lesson is part of the following series:

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

Taught by Professor Edward Burger, this lesson comes from a comprehensive Calculus course. 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'Hopital'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

2174 lessons

Founded in 1997, Thinkwell has succeeded in creating "next-generation" textbooks that help students learn and teachers teach. Capitalizing on the power of new technology, Thinkwell products prepare students more effectively for their coursework than any printed textbook can. Thinkwell has assembled a group of talented industry professionals who have shaped the company into the leading provider of technology-based textbooks. For more information about Thinkwell, please visit www.thinkwell.com or visit Thinkwell's Video Lesson Store at http://thinkwell.mindbites.com/.

Thinkwell lessons feature a star-studded cast of outstanding university professors: Edward Burger (Pre-Algebra through...


Recent Reviews

This lesson has not been reviewed.
Please purchase the lesson to review.
This lesson has not been reviewed.
Please purchase the lesson to review.

L'Hospital's Rule
Other Indeterminate Forms
Another Example of One to the Infinite Power Page [1 of 2]
I've got to tell you, this 1^[ ]camouflaged indeterminate form, I think, is my personal favorite. Not only can you come up with a way of thinking about e, but it really drives home these fundamental ideas about logarithms and exponentials and really how valuable using those things in harmony really is. And I'll tell you the thing that I find really cool is that you're faced with this really, really hard question. You've got some sort of complicated formula or some sort of complicated function with x's here, and then also x's in the exponent, and then somehow, by putting a log in front, it allows you to pull the exponent and make it now just a factor. And then you can use these old techniques. But you just can't put a log in front for free. There's no such thing as a free long. What you've got to do is you've got to put an exponential to kill off the log. So it's sort of a two-step process. You want the log, you pay for it with an exponential. Once you've got the log, you're allowed to pull this really complicated variable exponent out as a coefficient. Then you sort of work on this limit first, the exponent limit. And, at the very end of the day, remember that the answer will be e to that power. That really is a neat idea. So I wanted to show you one last one, just so we could do one together and really have fun with 1^, because you've just got to love it.
How about this - this limit, as x approaches 1, of . Now, as always, the first thing we want to do is see if we can just take the limit. So as x approaches 1, this approaches 1. So that's great, we like 1. But what about this? Well, as x approaches 1, the denominator is approaching zero, so this whole thing, unfortunately, is approaching infinity. So we're back to one of my favorite camouflaged indeterminate forms, the 1^ thing. So how do we make this into something that's more familiar, which will allow us to use L'Hôspital's Rule? Well, the answer is to use these two basic facts. Fact one is that if I take e^ln of junk, I get junk. So I could write it in a very complicated way and it doesn't change anything at all. Why would you possibly want to do that? Well, because of fact two, and fact two tells us that if you have natural log of something to a power, that power migrates out in front and becomes a coefficient. And the thing that makes this question really hard is all the x stuff that's up in the exponent. I want to pull that down. To pull that down, I need a log, but to make myself allow to write an equal sign, I need the e. So these two steps allow me to crack this question. So let's see them in action.
So the first step is to write this thing in a very, very complicated way. So I'm going to write it as . Now, this equal sign follows by this basic properties of logs and exponents. e^ln of junk equals junk, so e^ln of this thing equals that thing. So these are really equal. Okay, great, so now, why would I do that? Because now I can use the fundamental fact about logarithms, that this exponent, which is the thorn in our side, becomes a coefficient, which is a much nicer, friendlier thing. So now, I'm going to use this property to take this and pull this out in front. And if I pull this out in front, just in front of the log, what do they look like? Now, it would look like this. This would equal the limit as x approaches 1 of e^1/(1-x)ln x. So I pulled that out in front to here. Okay, well, now what do I do? Well, I look and say, "Yikes!" However, whatever this limit is, all I've got to do is look at the exponent thing. And if that answer turns out to be 15, then I know that this whole limit will be e^15. If this turns out to be 45, then I know this answer is e^45. So all I've got to do is do this little subproblem first, and then whatever that answer is, I know the answer to this question is going to be e to that. So, in fact, let me write that in here. e^? - who knows? This is the mystery. So let me go off and do a little subproblem.
And so the subquestion - there's no equal sign here. This limit does not equal the previous limit. It's only going to equal the exponent of the previous limit. So it's the limit, as x approaches 1, of 1/1(1x)ln x. Now, I'm going to write that in the way that we really probably should write it, . So this is the exponent. Let's take a look at what happens as we let x approach 1. The natural log of 1 = 0, so this is going to 0. And what about the bottom? Well, as x approaches 1, the bottom goes to 0. Finally, a friendly indeterminate form, a friendly face. So what I can see now is that the 0/0 indeterminate form I can conquer using none other than, you guessed it, L'Hôspital's Rule. So I finally hit something familiar. So let's use L'Hôspital's Rule to see what this indeterminate form equals. So it equals the limit as x approaches 1. And now, I take the derivative of the top and see 1/x divided by the derivative of the bottom. And the derivative of the bottom is just negative 1. And so what I see here is the limit, as x approaches 1, of negative -1/x. And that limit is negative 1.
So you're saying "Wow! We're done, we did it. The answer is negative 1," and if it were a multiple-choice test and you saw negative 1 as choice c, you would put it down. But that would be such a tragedy, because, after all of that great work, we forgot that we're working with that subliminal problem. This is just the limit of the exponent. So the limit of the exponent is negative 1. But remember the original question. The original question was this. So we just saw that this top is negative. So, in fact, the actual answer is this, which is another way of saying 1 over e. So looks how e makes its appearance in these questions. That's so neat. This limit turns out to equal 1 over e, 1 over 2.718 blah, blah, blah. So it's really neat. And the strategy, I think, is really neat just as well. The idea is you want a log to allow you to move this out in front, but to allow you to put an equal sign there, you have to put an e there. You do it like this and then you sort of ignore the e for a second and do a subliminal problem, look at the limit of the exponent, find the answer to that, come back to the liminal problem - I guess if this is a subliminal problem, this would be the liminal problem. You'd put it back into the liminal problem and you get the answer, 1/e.
Congratulations! You have conquered the indeterminate form! I'll see you at the next lecture.

Embed this video on your site

Copy and paste the following snippet: