Log in Victoria 11 years agoPosted 11 years ago. Direct link to Victoria's post “This is just more of a ge...” This is just more of a general question, but why doesn't infinity divided by infinity just equal 1? • (19 votes) Jeremy 11 years agoPosted 11 years ago. Direct link to Jeremy's post “Well, one reason is that ...” 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. (49 votes) Michael Berrios 10 years agoPosted 10 years ago. Direct link to Michael Berrios's post “at 1:04 when Sal says to...” at • (17 votes) Enn 10 years agoPosted 10 years ago. Direct link to Enn's post “*L'Hopital's* rule can be...” L'Hopital's rule can be applied in all cases of indeterminate forms which include 0/0, infinity / infinity, (infinity) /(-infinity) and Giuseppe 12 years agoPosted 12 years ago. Direct link to Giuseppe's post “Sal my teacher told me a ...” 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. • (13 votes) Marty Anderson 12 years agoPosted 12 years ago. Direct link to Marty Anderson's post “The first time he introdu...” The first time he introduced limits approaching infinity, that was the way he did it. (27 votes) Krzysztof Żurad 13 years agoPosted 13 years ago. Direct link to Krzysztof Żurad's post “How do we know, when are ...” How do we know, when are we supposed to use L'hopital's rule???? 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) Ivan Barreto 12 years agoPosted 12 years ago. Direct link to Ivan Barreto's post “L'hopital's rule is just ...” 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. (16 votes) Akbar Khuwaja 12 years agoPosted 12 years ago. Direct link to Akbar Khuwaja's post “Sal used L'Hopital's Rule...” 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) at Can somebody please confirm if my approach is correct? Thank you! • (7 votes) jbassettsea 12 years agoPosted 12 years ago. Direct link to jbassettsea's post “It is correct. You did wh...” 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) mohzakiyah1997 a year agoPosted a year ago. Direct link to mohzakiyah1997's post “another easy way to find ...” 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 • (10 votes) anemailyoucanremember 9 years agoPosted 9 years ago. Direct link to anemailyoucanremember's post “at 2:47 Sal says that "It...” at • (2 votes) Creeksider 9 years agoPosted 9 years ago. Direct link to Creeksider's post “At that point in the anal...” 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) Don Jaydee Emilio Tarpeh 12 years agoPosted 12 years ago. Direct link to Don Jaydee Emilio Tarpeh's post “Hey!! I am learning soooo...” 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) Hunter Thompson 10 years agoPosted 10 years ago. Direct link to Hunter Thompson's post “I see the relationship yo...” 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). (2 votes) scott albert 7 years agoPosted 7 years ago. Direct link to scott albert's post “Hi. This question may be ...” 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) Fai 7 years agoPosted 7 years ago. Direct link to Fai's post “You're applying L'Hopital...” 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) Daniele Pellegrini a year agoPosted a year ago. Direct link to Daniele Pellegrini's post “I would also have said th...” 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)Want to join the conversation?
1:04
(-infinity) / (infinity).
and when just normal limit?? 2:03 2:47
Video transcript
We need to evaluate the limit,as x approaches infinity, of 4x squared minus 5x, all of thatover 1 minus 3x squared. So infinity is kindof a strange number. You can't just plug in infinityand see what happens. But if you wanted to evaluatethis limit, what you might try to do is just evaluate-- if youwant to find the limit as this numerator approaches infinity,you put in really large numbers there, and you're going to seethat it approaches infinity. That the numeratorapproaches infinity as x approaches infinity. And if you put really largenumbers in the denominator, you're going to seethat that also-- well, not quite infinity. 3x squared will approachinfinity, but we're subtracting it. If you subtract infinity fromsome non-infinite number, it's going to be negative infinity. So if you were to just kindof evaluate it at infinity, the numerator, you wouldget positive infinity. The denominator, you wouldget negative infinity. So I'll write it like this. Negative infinity. And that's one of theindeterminate forms that L'Hopital's Rulecan be applied to. And you're probably saying,hey, Sal, why are we even using L'Hopital's Rule? I know how to do thiswithout L'Hopital's Rule. And you probablydo, or you should. And we'll do that in a second. But I just wanted to show youthat L'Hopital's Rule also works for this type of problem,and I really just wanted to show you an example that had ainfinity over negative or positive infinityindeterminate form. But let's applyL'Hopital's Rule here. So if this limit exists, or ifthe limit of their derivatives exist, then this limit's goingto be equal to the limit as x approaches infinity of thederivative of the numerator. So the derivative of thenumerator is-- the derivative of 4x squared is 8x minus 5over-- the derivative of the denominator is, well,derivative of 1 is 0. Derivative of negative 3xsquared is negative 6x. And once again, when youevaluated infinity, the numerator is going toapproach infinity. And the denominator isapproaching negative infinity. Negative 6 times infinityis negative infinity. So this is negative infinity. So let's applyL'Hopital's Rule again. So if the limit of these guys'derivatives exist-- or the rational function of thederivative of this guy divided by the derivative of that guy--if that exists, then this limit's going to be equal tothe limit as x approaches infinity of-- arbitrarilyswitch colors-- derivative of 8x minus 5 is just 8. Derivative of negative6x is negative 6. And this is just going to be--this is just a constant here. So it doesn't matter what limityou're approaching, this is just going to equal this value. Which is what? If we put it in lowestcommon form, or simplified form, it's negative 4/3. So this limit exists. This was an indeterminate form. And the limit of thisfunction'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 whosay, hey, we already knew how to do this. We could just factorout an x squared. You are absolutely right. And I'll show youthat right here. Just to show you that it'snot the only-- you know, L'Hopital's Rule is notthe only game in town. And frankly, for this type ofproblem, my first reaction probably wouldn't have been touse L'Hopital's Rule first. You could have said that thatfirst limit-- so the limit as x approaches infinity of 4xsquared minus 5x over 1 minus 3x squared is equal to thelimit as x approaches infinity. Let me draw a little line here,to show you that this is equal to that, not to thisthing over here. This is equal to the limitas x approaches infinity. Let's factor out an x squaredout of the numerator and the denominator. So you have an x squaredtimes 4 minus 5 over x. Right? x squared times 5over x is going to be 5x. Divided by-- let's factor outan x out of the numerator. So x squared times 1over x squared minus 3. And then these xsquareds cancel out. So this is going to be equalto the limit as x approaches infinity of 4 minus 5 over xover 1 over x squared minus 3. And what's this goingto be equal to? Well, as x approachesinfinity-- 5 divided by infinity-- this termis going to be 0. Super duper infinitelylarge denominator, this is going to be 0. That is going to approach 0. And same argument. This right here isgoing to approach 0. All you're left with isa 4 and a negative 3. So this is going to beequal to negative 4 over a negative 3, or negative 4/3. So you didn't have to douse L'Hopital's Rule for this problem.
