AP®︎ Calculus BC (2017 edition)
Sal uses L'Hôpital's rule to find the limit at infinity of (4x²-5x)/(1-3x²). Created by Sal Khan.
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- This is just more of a general question, but why doesn't infinity divided by infinity just equal 1?(19 votes)
- Well, one reason is that two quantities could both approach infinity, but not at the same rate. For example imagine the limit of (n+1)/n^2 as n approaches infinity. Both the numerator and the denominator approach infinity, but the denominator approaches infinity much faster than the numerator. So take a very large n, like 1 trillion. The numerator is 1,000,000,000,001. But the denominator is 1 trillion SQUARED. So (n+1)/n^2 for n=1 trillion is .000000000001000000000001. That's very close to zero, not 1. As n gets even bigger, the limit of (n+1)/n^2 approaches even closer to 0.(45 votes)
- at1:04when Sal says to evaluate for infinity, does it matter if it comes out as infinity/infinity or pos. infinity/ neg. infinity in order for l'hopitals rule to still apply?(16 votes)
- L'Hopital's rule can be applied in all cases of indeterminate forms which include 0/0, infinity / infinity, (infinity) /(-infinity) and
(-infinity) / (infinity).(33 votes)
- Sal my teacher told me a short cut. If f(x) is a approaching a infinity and the powers are the same (ex. 4x^2/3x^2)then the coefficients become what the limit approaches. Without even factoring you can tell that the answer is going to be 4/3. Does that work all the time?
Also thank you. I learned someone awesome in this video. :) As usual.(12 votes)
- How do we know, when are we supposed to use L'hopital's rule????
and when just normal limit??
How many times do i have to make :'hopital's rule, to get the result?? (in this harder cases?) Coz i can just make again and again, and what if i dont get the result?(10 votes)
- L'hopital's rule is just one of many different methods to get the result for a limit. If you see you are not getting results with it, try another approach. There is no way to tell the best way to solve an limit, that's just trial and error.(15 votes)
- Sal used L'Hopital's Rule twice. I solved the question by using L'Hopital's Rule once.
When I apply the limit (as x approaches infinity) to the derivative (8x-5) / (-6x) at2:03, I know that "-5" is negligible when x reaches infinitely large values. That leaves (8x) / (-6x), which is scaling at the rate of 8 / -6, or 4 / -3. Hence, the answer to the question asked.
Can somebody please confirm if my approach is correct? Thank you!(7 votes)
- It is correct. You did what Sal said at the begining of the vidieo about already knowing how to get the limit. While your way is correct, Sal only continued taking L'Hopital's rule one step further to get the same result.(7 votes)
- another easy way to find the limit is to take the number beside high the variables with the highest power. 4x^2/-3x^2==== -4/3(9 votes)
- at2:47Sal says that "It doesn't matter what value this [limit] is approaching." I'm not sure I understand how that could be? The fraction would not be there at any point besides infinity, because otherwise it wouldn't be indeterminate, to my thinking. Or am I just reading too much into it? Thanks.(2 votes)
- At that point in the analysis, we're looking at the limit of a constant as x goes to infinity, and Sal is making the point that a constant will always have the same value no matter what limit we're approaching. He isn't saying that the specific limit (infinity) is irrelevant to the overall problem, but just noting that once you've got a constant you're done and don't have to worry about the limit any more.(10 votes)
- Hey!! I am learning sooooo much more from these presentations than I do in my lectures.... can we also us this rule to find the vertical and horizontal asymptotes? how do we use it from the + and - sides?(4 votes)
- I see the relationship you're foreseeing, since both deal with the slope of functions where values zero and infinity occur. For horizontal asymptotes, I suspect you may be right, that L'Hopital's Rule can help us find the value f(x) for which the slope approaches 0.
However, I don't believe the two are related when it comes to vertical asymptotes, since these deal with slopes of undefined forms (e.g., 5/0 or the general form n/0), not of indeterminate form (i.e., 0/0 or infinity / infinity).
Hope this helps!
Also, if you'd like a general review of horizontal and vertical asymptotes, you can watch the video below (although it's rather long, at 20 minutes, I'm sure you can parse through it).
- Hi. This question may be way off, but when we are calculating the derivative for (4x^2-5x)/(1-3x^2), why are we not using the quotient rule?(2 votes)
- You're applying L'Hopital's rule to calculate the limit at infinity. You're not interested in the derivative of this function (in which case, yes, the quotient rule can be used).(6 votes)
- I would also have said that when x approaches to infinity, the two quadratic terms go to infinity much faster than the linear term. So -5x will become negligible compared to 4x^2, 1 of course is negligible and so we have 4x^2/(-3x^2) which is exactly -4/3.(4 votes)
We need to evaluate the limit, as x approaches infinity, of 4x squared minus 5x, all of that over 1 minus 3x squared. So infinity is kind of a strange number. You can't just plug in infinity and see what happens. But if you wanted to evaluate this limit, what you might try to do is just evaluate-- if you want to find the limit as this numerator approaches infinity, you put in really large numbers there, and you're going to see that it approaches infinity. That the numerator approaches infinity as x approaches infinity. And if you put really large numbers in the denominator, you're going to see that that also-- well, not quite infinity. 3x squared will approach infinity, but we're subtracting it. If you subtract infinity from some non-infinite number, it's going to be negative infinity. So if you were to just kind of evaluate it at infinity, the numerator, you would get positive infinity. The denominator, you would get negative infinity. So I'll write it like this. Negative infinity. And that's one of the indeterminate forms that L'Hopital's Rule can be applied to. And you're probably saying, hey, Sal, why are we even using L'Hopital's Rule? I know how to do this without L'Hopital's Rule. And you probably do, or you should. And we'll do that in a second. But I just wanted to show you that L'Hopital's Rule also works for this type of problem, and I really just wanted to show you an example that had a infinity over negative or positive infinity indeterminate form. But let's apply L'Hopital's Rule here. So if this limit exists, or if the limit of their derivatives exist, then this limit's going to be equal to the limit as x approaches infinity of the derivative of the numerator. So the derivative of the numerator is-- the derivative of 4x squared is 8x minus 5 over-- the derivative of the denominator is, well, derivative of 1 is 0. Derivative of negative 3x squared is negative 6x. And once again, when you evaluated infinity, the numerator is going to approach infinity. And the denominator is approaching negative infinity. Negative 6 times infinity is negative infinity. So this is negative infinity. So let's apply L'Hopital's Rule again. So if the limit of these guys' derivatives exist-- or the rational function of the derivative of this guy divided by the derivative of that guy-- if that exists, then this limit's going to be equal to the limit as x approaches infinity of-- arbitrarily switch colors-- derivative of 8x minus 5 is just 8. Derivative of negative 6x is negative 6. And this is just going to be-- this is just a constant here. So it doesn't matter what limit you're approaching, this is just going to equal this value. Which is what? If we put it in lowest common form, or simplified form, it's negative 4/3. So this limit exists. This was an indeterminate form. And the limit of this function's derivative over this function's derivative exists, so this limit must also equal negative 4/3. And by that same argument, that limit also must be equal to negative 4/3. And for those of you who say, hey, we already knew how to do this. We could just factor out an x squared. You are absolutely right. And I'll show you that right here. Just to show you that it's not the only-- you know, L'Hopital's Rule is not the only game in town. And frankly, for this type of problem, my first reaction probably wouldn't have been to use L'Hopital's Rule first. You could have said that that first limit-- so the limit as x approaches infinity of 4x squared minus 5x over 1 minus 3x squared is equal to the limit as x approaches infinity. Let me draw a little line here, to show you that this is equal to that, not to this thing over here. This is equal to the limit as x approaches infinity. Let's factor out an x squared out of the numerator and the denominator. So you have an x squared times 4 minus 5 over x. Right? x squared times 5 over x is going to be 5x. Divided by-- let's factor out an x out of the numerator. So x squared times 1 over x squared minus 3. And then these x squareds cancel out. So this is going to be equal to the limit as x approaches infinity of 4 minus 5 over x over 1 over x squared minus 3. And what's this going to be equal to? Well, as x approaches infinity-- 5 divided by infinity-- this term is going to be 0. Super duper infinitely large denominator, this is going to be 0. That is going to approach 0. And same argument. This right here is going to approach 0. All you're left with is a 4 and a negative 3. So this is going to be equal to negative 4 over a negative 3, or negative 4/3. So you didn't have to do use L'Hopital's Rule for this problem.