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Calculus: Exotic Examples of Indeterminate Forms


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  • Type: Video Tutorial
  • Length: 12:48
  • Media: Video/mp4
  • Use: Watch Online & Download
  • Access Period: Unrestricted
  • Download: MP4 (iPod compatible)
  • Size: 138 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: Indeterminate Quotients (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 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.

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L'Hospital's Rule
Indeterminate Quotients
More Exotic Examples of Indeterminate Forms Page [1 of 3]
So now let's take a look at some more exotic examples of indeterminate forms. I want to start with what I refer to as the classics. These are the ones that every math person just loves to look at because, in fact, these limits come up a lot and old-fashioned methods that we learned about a long time ago in factoring and stuff won't cut it with these classics. Here we go. Let's take a look at some classics. The first one I want to take a look at is the following. The limit as x approaches 0 of sin x, a trigonometric function, over x. Now let's take a look at this. As x approaches 0, the very first thing I always do is just try to take the limit and see what happens. If x approaches 0, well sin x, that's going to actually approach 0. Sin 0 is 0. In fact, do you remember that? Here's how I think of that by the way. This has nothing to do with anything, but this is just between us. I always just visualize. In my mind's eye I just see the sin function, be the sin function.
There it is. So at 0 it's 0. So, it's 0 all right. Now, what about on the bottom? Well, the bottom is approaching 0 too. So, this is ^0/[0]. So this is indeed an indeterminate form and it's of the flavor ^0/[0]. What to do? Well, I'm going to apply L'Hospital's rule. So we apply L'Hospital's rule, all I've got to do is take the derivative of the top, derivative of the bottom and see what that new thing equals. So here we go. So we take the limit x approaches 0 and now I take the derivative of sin, well that's cos x, divided by the derivative of x, well that's just 1. So, what's the limit of this? Well, as x approaches 0, cos x approaches 1. So, in fact, this is approaching ^1/[1], which is 1. So, the moral of the story is, the limit as x approaches 0 of sin x over x is one. Neat.
Okay, great, let me show you another example. In fact, I'll just do it right here. Limit as x approaches 0, well, if sin has its day, then plainly we need to see cosine. So where's cosine? Here it is. Classic example number two. Cos x - 1 all over x. What's the limit here? Well, as x approaches 0, cos x approaches 1. So 1 - 1 is 0, ^0/[0], that's an indeterminate form of the flavor ^0/[0]. So, I'm going to use L'Hospital's rule. So to use L'Hospital's rule I take the derivative of the top and the derivative of the bottom. Derivative of the top, derivative of cos is -sin x and the derivative of the bottom is 1. What's the limit as x goes to 0 of this? Well, on the top I see -sin 0, which is -0, which is 0. 0 on top, 1 on the bottom, not an indeterminate form. It equals 0. Right? ^0/[1] is 0. So, moral of the story is that the limit as x approaches 0 of cos x - 1 over x, that limit is 0. Two classic examples.
Well let's now move away from the classics and just look at some more real exotic ones. Here's some really exotic ones. This is great. You can show your friends this, they'll be really, really impressed. So the first one of the exotic genre that is not classic. These are like modern. It's sort of like funk, you know. These are the funk limits. Sin x over e^x. Let's see what happens here. Sin x over e^x. Okay, so we can start to take derivatives if you want, but the first thing I always ask myself is, "Can I do it?" Now if x approaches 0, sin x approaches 0, but what is e^x approaching? If x is approaching 0, this is approaching e^0 and e^0 is actually 1. So, in fact, this is approaching ^0/[1], which is not an indeterminate form. No, no, no, it's not indeterminate form. It's in fact, ^0/[1], which equals 0. So, in fact, this limit can be done without applying L'Hospital's rule. It would be a very, very sad thing if you actually applied L'Hospital's rule. You'd get the wrong answer. So again, always start off by seeing if you need to actually use L'Hospital's rule at all.
So here's an example. Here's another one that's funky for a different reason. You'll see why. First of all, let's let x race off to infinity and then I'll have x^2 + 3 all over x ln x. So there's the funky part there. The funky part there is that I've got the natural log. So, that's looking a little scary. Now, what's going on here? As goes off to infinity, the top is racing off to infinity. Right? Then what's x ln x? Well, x and ln x are both racing off to infinity. So this whole thing is racing off to infinity. Right? So, what do you do? Well, it's an indeterminate form of the formula infinity over infinity.
Let's make a guess as to what the answer is. Now what would be a good guess? Well, let's see. You see, I'll look at this and I see this thing is growing like x²; the +3 is insignificant. So this thing is really and x². What's going on on the bottom here? Well, this is x times the ln x. So the x and the x up here, they sort of match up, but what about the extra x, because I have two copies of x here, against the natural log? What grows faster? Well let me just tell you how I think about this. This is not required for using L'Hospital's rule, but just to show you that you can actually think about these things a little bit and then see if you can make a good guess as to the answer.
I happen to remember what the natural log functions looks like. The natural log graph looks like that. Whereas just the x graph, y = x looks like this. This is x and this is the natural log. So which is growing faster? Well plainly the x function is growing faster. So while this x and one of these x's sort of kill each other off, I'm left with an extra x and a natural log. Who would win that race? It looks like the x is growing much faster than the natural log. Which would seem to indicate that the top is growing faster than the bottom. If the top is racing off to infinity, faster and faster and faster than the bottom, then that would seem to indicate this whole thing should go to infinity, because the top is going to infinity faster than the bottom.
So, let's see if that guess is right. Now, if you don't have a guess, that's okay, but I'm just pointing out the fact that sometimes you can look at these things and at least make an educated guess and see if that's right. Let's see if we can use L'Hospital's rule to either verify or not, my guess. So, I take the derivative of the top, which is just 2x and on the bottom, I have to take the derivative of that. Now that actually requires a product rule. The dominator is a product. Its x ln x. So the product rule says the first times the derivative of the second plus the second times the derivative of the first. So here we go. The first times the derivative of natural log, which is ^1/[x], plus the second, that's natural log of x, times the derivative of the first, and that's just 1. So, what do I see? Well, I see that this equals--so what does this equal? Well, I have the limit as x approaches infinity of, and I have a 2x on top, divided by--and the x divided by x, they actually cancel, so I'm left with just a 1 + ln x.
What's the limit as x goes to infinity now? Well this is still going to infinity and the bottom is still going to infinity. This remains an indeterminate form of the flavor infinity over infinity. So, it needs more work. What will I do? I'll apply another application of L'Hospital's rule. L'Hospital's rule is sort of like potato chips. You can't just have one. So, on the top the derivative would be just 2. On the bottom, what's the derivative? The derivative of 1 is 0. The derivative of the natural log is ^1/[x]. This is going to be a little tricky, we're going to have a double flip here. So, what does that equal? Well, that equals the limit as x goes to infinity of--if I flip this, invert and multiply, it comes up on top, so I see 2x, and what's that limit? Well, as x approaches infinity, this goes to infinity. Which is exactly our prediction early on when I took a look and said the top is actually growing faster than the bottom. So this limit is actually infinite or some might say doesn't exist. It just is not a number.
How about one last one just for fun? Because I know once you start, it's so hard to stop. How about the limit as x approaches ^? Okay, so remember this is ^of the tangent 2x divided by the tangent of 2x. So I'm throwing in a little trig action here. Let's see how we do with a little trig action. Well the first thing I do is see if I can take the limit. If I take the limit I see, well if x is approaching ^^then this term right here is approaching 2(^), which is just . So - is 0. That's okay. But what about tangent of--and if I plug this in I get --what's tangent of ? Well tangent of is also 0. So, this is ^0/[0]. It's an indeterminate form of the variety ^0/[0].
So, let's us a little L'Hospital's see if that sets us on the right path. We're still taking the limit as x approaches ^but now I take the derivative of the top. The derivative of the top, that's just a constant, so the derivative of that is 0. So just see -2 on top and the bottom, what do I see? Well this is actually a little chain rule. I've got to use a little chain rule right? I've got an inside thing. Right? I've got the 2x and then I've got the outside thing. Right? Tangent. So, I've got tangent of blob. So, what's the derivative of tangent of blob? Derivative of tangent of blob is sec² of blob and the blob is 2x. I have to multiply that by the derivative of the blob and the derivative of the blob is just 2. Put the 2 out in front, so I see that. So, what do I see? Well the 2's cancel, so that's sort of nice. Left with a -1 on top. So, I'm left with the limit as x approaches ^. I've still got the -1 on the top. On the bottom I have sec²(2x).
Okay, now I've got to take the limit. Now some of you are just fans of secant. Those of you who are, right on. I unfortunately am not, because secant, who knows. But I do remember one fact about secant. Secant is 1 over cosine. So, this thing right here is 1 over cos² of 2x. So if I invert and multiply, so this is -1 over 1 over cosine. If I inverse and multiply I see a negative--oh I see the limit of course, always see the limit, x approaches ^and then I see -cos² (2x). Now cos, I'm a little more comfortable thinking about. So what is--let's see, if x approaches ^then this is approaching and what's cosine of ? Well cosine of is -1 and -1² would be 1 and that negative sign in front gives me a -1. So there's the answer.
Okay, so I hope you get a chance to sort of think about this L'Hospital's rule stuff and see really that it empowers us to take limits of really exotic things that first appear to be indeterminate. Okay, see you next.

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