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# Limit at infinity of a difference of functions

Sal finds the limit at infinity of √(100+x)-√(x). Created by Sal Khan.

## Want to join the conversation?

• At when 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.
• 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?
• 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.
• Could you have intuitively taken 100 out and then at that point subtracted root x by root x and gotten 0?
• 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.
• at - Sal 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.
• 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")?
• 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.
• 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
• 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.
• 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.
• 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?
• 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.