Calculus, all content (2017 edition)
Course: Calculus, all content (2017 edition) > Unit 1Lesson 20: Limits at infinity (horizontal asymptotes)
- Infinite limits intro
- Limits at infinity of quotients (Part 1)
- Limits at infinity of quotients (Part 2)
- Limits at infinity of quotients
- Limits at infinity of quotients with square roots (odd power)
- Limits at infinity of quotients with square roots (even power)
- Limits at infinity of quotients with square roots
- Limits at infinity of quotients with trig
- Limits at infinity of quotients with trig (limit undefined)
- Limits at infinity of quotients with trig
- Limit at infinity of a difference of functions
Limit at infinity of a difference of functions
Sal finds the limit at infinity of √(100+x)-√(x). Created by Sal Khan.
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- At5:00when you've simplified the limit to 100/sqrt(100+x)+sqrt(x), would it be possible to further simplify? My brain instantly wants to pull the 100 out of the bottom sqrt and end up with 100/10sqrt(x) + sqrt(x) then simplify to 10/sqrt(x) + sqrt(x) then 10/2sqrt(x) then finally 5/sqrt(x). I feel like I'm making a mathematical mistake, but I don't know where.(10 votes)
- I understand multiplying by the conjugate to allow simplification via difference of squares, but I got stuck with the questions that needed both the numerator and denominator being divided by x, and then turning an x into an x^2 to get it inside the radical. None of this was intuitively obvious to me at first, and I'm wondering how many more intuitive 'tricks' like this are required to solve common textbook and exam questions. I feel like I need an 'algebra manipulation tricks' playlist. Or have I just missed the videos where Sal explains this stuff?(3 votes)
- BTW second question of this video exactly uses what Kevin is referring to: https://www.khanacademy.org/math/calculus/limits_topic/old-limits-tutorial/v/limit-examples-w--brain-malfunction-on-first-prob--part-4
As to directly answering your question, sorry I don't have a great answer. Perhaps you might get a relief that people other than you (at least me) are wondering about this too. I'm frustrated as well, wondering how many "tricks" do I need to "memorize." I'm by no means a genius, so I can't spontaneously come up with these "tricks." But I understand them when I see the explanations and can utilize them in similar questions in the future. I think this will come naturally, as we practice more and more questions. I also would like some "trick playlist" that covers every trick that we need to know to solve ANY question, but I'm sure there's none like that.
Simply put, more practice! is the answer.(5 votes)
- at3:06-3:44Sal simplifies to (Sqrt100 - X)(Sqrt100 + X)/(Sqrt 100 + X). Why not cancel out and just have (Sqrt100 - X)(1)?(1 vote)
- Think about what Sal is trying to do here. He is trying to get rid of the square roots in his expression. If you cancel out to get sqrt(100-x), then you've just gone back to where you started, no closer to the answer.(6 votes)
- Can limits at positive or negative infinity be determined algebraically? Up to this point it seems that it's mostly just figuring out what "looks right" (or as Sal calls it: "hand waving") as some variable approaches infinity. That method is all fine and dandy, but what if you have to find the limit of some very complex function as it approaches infinity, the method used up to this point would seem to be impractical or inaccurate. Plus from past experience I know that when you're taught to do something "by hand" in math there's some more abstract and simpler way of doing it.
So, in short, is there some more abstract higher level method of determining limits at infinity than the current one (which seems just like choosing what "looks right")?(4 votes)
- In this particular video I didn't see any hand-waving. Sal presented a typical algebraic manipulation of an expression into an equavalent expression which is easier to determine the limit. These types of algebraic manipulation become VERY important in later math, it is not just some trick that worked in this case. Integration and proofs often rely on this technique in order to be solved.
Having said that, the limit of the original question as it stood, could have just as easily been shown to be zero. In less than a year, you will be able to look at many of the examples used here and and know what the limit is without having to calculate it because you will then be familiar with the behavior of various types of expressions as they tend to infinity, as well as becoming more comfortable with the concept of infinity itself.
Also, later on, you will be learning much more analytical methods regarding limits and convergence in particular (in the example, the limit as x approaches infinity converges to the value 0). For example, the Cauchy Sequence. If you can prove that the elements of the sequence become arbitrarily close to each other as the sequence progresses, then you know it has a limit, even if you don't know what the limit is. This is just one of many modes of convergence. Search the phrase "modes of convergence" to find out more.(0 votes)
- I am i little confused so for the example sqaure root (9x^2+4x ) - 3x . I intially though that in the radical sign 9x^2 would dominate and outside 3x would dominate leaving the 4x inside negligble . Then you are left with 3x-3x essentially which would give you an answer of 0 as the limit approached infinity but this is not the case and I am not sure why(2 votes)
- The 4x isn't negligible. To see why, start out tackling this problem the same way Sal handles the problem in this video. After multiplying by the conjugate and simplifying, you get a numerator of 4x, and a denominator of sqrt(9x^2+4x) + 3x. At this point you can see why the 4x is important, because it leaves a numerator of the same order as the denominator. Now you can divide through by x. In the numerator you get 4, and in the denominator you get sqrt(9+4/x) + 3. So now we can see that as x goes to infinity we get 4 divided by sqrt(9) + 3, which is 4/6 = 2/3.(3 votes)
- I still don't understand the need to algebraically manipulate a defined function. Help, please?(1 vote)
- When taking limit, we often find ourselves in an indeterminate form, such as a/0, 0/0, or 0^0 and other forms of indeterminates, that we can't calculate. So the trick/technique is algebraic manipulation. By manipulating it, we can turn it into something we can calculate.
For example, find the limit as x->1 of (x^2-1)/(x-1). If you try to plug in x = 1, you get 0/0, which is an indeterminate form. We can manipulate it by factoring and canceling
limit_x->1 (x^2 - 1)/(x-1) =
limit_x->1 (x+1)(x-1)/(x-1) =
limit_x->1 (x+1) = 2
As you see, by manipulating it algebraically, we can plug in x=1 and calculate as oppose to the original expression.(3 votes)
- Why (sqrt(100+x)+sqrt(x))/(sqrt(100+x)+sqrt(x))=1? What if x = 0? Or is this equation only equal to one when x->inf?(0 votes)
- Is the limit of a constant number undefined? I was wondering if, at the beginning of the video, the expression could be simplified as 10+sqrt(x) -sqrt(x), since 10 is the square root of 100, which would just leave 10.(1 vote)
- The limit of a constant function does exist. For example, the limit of f(x) = sqrt(100) would be equal to 10 for any value of x.
The difference in this expression is that x is added to 100 before the square root is taken. If you think about it, the square root of 100 plus the square root of x is not necessarily equal to the square root of 100 + x. For example, sqrt(100) + sqrt(25) = 15, while sqrt(100 + 25) = sqrt (125) which approximately equals 11. That's why we can't just simplify to 10. I hope this helps.(3 votes)
- At5:14, instead of multiplying out by the conjugate why couldn't we expand the original function to sqrt(100) + sqrt(x) - sqrt(x)? If we did it this way the two sqrt(x) 's would cancel and you would be left with sqrt(100) which equals 10.(1 vote)
- That would be because:
√(a + b)does NOT equal
√a + √b
√(100 + x) - √xis NOT equal to
√100 + √x - √x
You may need to review the properties of powers and roots as they are prerequisite to studying calculus.(2 votes)
- Could you have intuitively taken 100 out and then at that point subtracted root x by root x and gotten 0?(0 votes)
- No. You'd get the right answer, but for the wrong reasons. You need to be able to logically walk through the problem and show that the answer makes sense.(4 votes)
- Let's think about the limit of the square root of 100 + x - the square root of x, as x approaches infinity and I encourage you to pause this video and try to figure this out on your own. So I'm assuming you've had a go at it. So first let's just try to think about it, before we try to manipulate this algebraically, in some way. So what happens is x gets really really really really large, as x approaches, as x approaches infinity. While even though this 100 is a reasonably a reasonably large number, as x gets really large, Billion, trillion, trillion trillions. Even larger than that. Trillion trillion trillion trillions. You could imagine that the 100, under the radical sign, starts to matter a lot less. As x approaches really really large numbers, the square root of 100 + x is going to be approximately the same thing as the square root of x. So for really really large large x's, we can reason that the square root of 100 + x is going to be approximately equal to the square root of x and so in that reality, we are going to really really really large x's. In fact, there's nothing larger, where you can keep increasing x's, then these two things are going to be roughly equal to each other. So it's reasonable to believe, that the limit as x approaches infinity here, is going to be zero. You're subtracting this, from something that is pretty similar to that, but let's actually do some algrebraic manipulation to feel better about that, instead of this kind of hand-wavy argument about the 100 not mattering as much, when x gets really really really large and so let me re-write this expression. See if we can manipulate it in interesting ways. So this is 100 + x - x. So one thing that might jump out at you whenever you see one radical, minus another radical like this. Well maybe we can multiply by its conjugate and somehow get rid of the radicals, or at least transform the expression in some way, that might be a little more useful when we try to find the limit, as x approaches infinity. So let's just and obviously we can't just multiply it by anything arbitrary, in order to not change the value of this expression, we can only multiply it by one. So let's multiply it by a form of one, but a form of one that helps us, that is essentially made up of its conjugate. So let's multiply this let's multiply this times the square root of 100 + x + the square root of x, over the same thing. The square root of 100 + x + the square root of x. Now notice this, of course is exactly equal to one and the reason why we like to multiply by conjugates, is that we can take advantage of differences of squares. So this is going to be equal to and our denominator, we're just going to have we're just going to have the square root of 100. Let me write it this way actually. 100 + x + the square root of x and our then our numberator, we have the square root of 100 + x, minus the square root of x times this thing. Times square root of 100 + x + the square root of x. Now right over here, we're essentially multiplying A + B times A - B, will produce a difference of squares. So this is going to be equal to this top part right over here, is going to be equal to is going to be equal to this, let me do this in a different color. It's going to be equal to this thing squared minus, minus, this thing, minus that thing squared. So what's 100 + x squared? Well that's just 100 + x, 100 + x and then what square root of x squared? Well that's just going to be x. So minus x and we do see that this is starting to simplify nicely. All of that, over the square root of 100 + x + the square root of x and these x's x - x will just be nothing and so we are left with 100 over the square root of 100 + x + the square root of x. So we could re-write the original limit, as the limit, the limit as x approaches infinity. Instead of this, we just algebraically manipulated it, to be this. So the limit as x approaches infinity of 100 over the square root of 100 + x + the square root of x and now it becomes much clearer. We have a fixed numerator. This numerator just stays at 100, but our denominator right over here, is just going to be, it's just going to keep increasing. It's going to be unbounded. So if you're just increasing this denominator, while you keep the numerator fixed, you essentially have a fixed numerator, with an ever-increasing, or a super large, or an infinitely large denominator. So that is going to approach, that is going to approach zero, which is consistent with our original intuition.