Main content
AP®︎ Calculus AB (2017 edition)
Unit 1: Lesson 11
Analyzing limits at infinity- 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
© 2023 Khan AcademyTerms of usePrivacy PolicyCookie Notice
Limits at infinity of quotients with square roots (even power)
AP.CALC:
LIM‑2 (EU)
, LIM‑2.D (LO)
, LIM‑2.D.3 (EK)
, LIM‑2.D.4 (EK)
, LIM‑2.D.5 (EK)
Limit at infinity of rational expression with radical.
Want to join the conversation?
I had trouble while solving the practice exercise for this video. I kept making mistakes with the sign of my answers... meaning I knew the answer was 2 but positive or negative i just guessed based on whether infinity was positive or negative along with whether there were positive or negative terms in the numerator and denominator. Help Please.(55 votes)- I got it! I got it! if it's even power then the answer always be positive, no matter negative infinity or positive infinity. because negative * negative = positive.
I had the same trouble a few minuts ago, but everything is clear now.(42 votes)
At1:53, can someone please explain to me in more detail on how Sal simplified the radical expression in white?(14 votes)
Because x² will always be positive, so we can rewrite it as √x⁴. But you are probably confused how we go from x² to √x⁴. Remember that squaring a number and its square root undo each other, just like division and multiplication or subtraction and addition.
For example, we have a number x.
If we subtract 7 from x, we get x-7, but this is not x anymore. To make it x again, we need to undo what did by adding 7. So x - 7 + 7 would be equivalent to x.
Similarly if we do x/7 then we must undo it by multiplying by 7, and we get (x/7)*7
Now, if we square x, we get x² and to undo that we square root it, √x². But notice why I had stated that "because x² will always be positive". This is because when you square root , you get a positive and a negative number. Has it been -x², then it would be a negative number, and we would rewrite it as -√x⁴.
Back to how we go from x² to √x⁴. We start with x². If we square it (x²)², we must undo it by square root it, √((x²)²). Simplify it we get √x⁴. The next steps Sal didn't write it but I feel that I should to make it clear for you to understand. We can do similar process to the numerator to rewrite 1 = √1. So, 1/x² = √1 / √x⁴. By the radical properties, √1 / √x⁴ = √(1/x⁴). And again by the radical properties, Sal multiplied √(1/x⁴)√(4x⁴-x) together to get √((4x⁴-x)/x⁴) = √((4x⁴/x⁴)-(x/x⁴)) = √((4-(1/x³))
Hope that helps.(33 votes)
For people struggling to understand when to apply the negative sign, I just got it. For positive infinity, it doesn't matter. For negative infinity, think of it this way:
For any negative number, x to an odd power e.g. x^3 will result in a negative number because if x= -1, then -1*-1*-1 = -1.
This also applies for negative infinity. So as x approaches infinity, the result of x raised to any odd power should be negative (i.e. negative infinity).
But! If you're taking the square root of an even-numbered power, like when you do sqrt(1/x^6), that will make a POSITIVE number. So if you want that to be equivalent to 1/x^3, you can't just do sqrt(1/x^6), they are not equal!! (They are the same value but different sign).
So, instead you have to add the negative sign AFTER the square root operation, to make 1/x^3 = -sqrt(1/x^6). Hope that clears it up because I was very confused too.
TLDR, ONLY when there is a negative value for x (e.g. x approaches negative infinity) AND the power it is raised to is odd (e.g. x^3, x^5, etc.) then you need to add the sign to the side of the fraction with the square root.(27 votes)
For the past "limit at infinity" videos, Sal has just explained to find the dominating values and simplify them. Is this not the case anymore? Do/should we always do problems like this?(11 votes)- In previous video, Sal said "I'm not this in an ultra-rigous way but more in intuitive way". So in this video, he shows us the ultra- rigorous way.(18 votes)
- This question doesn't pertain to anything in this video, however my question is that when dealing with limits such as these, how do you determine whether your answer will be positive or negative. It's easy enough when approaching positive infinity, but when approaching negative infinity, sometimes the answer will be negative, so my question is how can you tell whether the answer is positive or negative?(12 votes)
For example, if you need to find the limit of the (square root of 4x^6) over (2x^3) at negative infinity, you would factor out a (negative square root of x^6) from the numerator, because x is going negative, not positive. That limit described above will be equal to -1, not 1.(2 votes)
- at2:20how the heck does (4x^4-x)/(x^4) simplify to 4 - (1/x^3) ?? I'm missing something, that for sure.(5 votes)
Recall that in general, (a - b)/c is equivalent to a/c - b/c.
So (4x^4-x)/(x^4) = (4x^4)/(x^4) - x/(x^4) = 4 - 1/(x^3).
If you are rusty with algebra, you will need to review algebra in order to have a realistic chance of performing well in calculus.(10 votes)
- Instead of doing this whole thing of multiplying and dividing by 1/x^2 why not just use the direct approach and take the highest degree terms. It is going to give us square root of 4x^4 divided by 2x^2. when we simplify this we get 1. So what exactly is wrong with this approach?(2 votes)
- I don't think there is anything wrong with this method. I think Sal maybe showing us a different way, using more algebra, to solve the limit.(15 votes)
- Hello everyone. I've got a question. In this video we multiply function by 1/x^2 at0:50, although x aproaches (-infinity) so should not we multiply this function by -1/x^2 instead. Because in a practice section i met a lot of similar tasks and sometimes i do not understand by what i should multiply, by "-" or by "+".(5 votes)
The function that Sal uses, namely 1/x^2, is independent of the value that x is approaching. If you did use -1/x^2 instead, you'd have to switch the signs of all the numbers in the expression, which is just more work, and another way to potentially get the wrong answer. Multiplying by a positive or negative expression depends on the case. Sometimes it makes things better and sometimes it makes it worse. With enough practice, you'll be able to tell more easily when to use the positive or the negative.(5 votes)
- At0:45, I still don't follow how you knew to multiply the numerator and denominator by 1/x^2. How am I supposed to know in future problems where to start with something like this?(4 votes)
So why did he chose this certain method. Is there a proof as to why this works? Are there more methods??(3 votes)- You could also solve this using L'Hopital's rule, but I wouldn't recommend it in this case since it'd quickly get "messy." The method shown in the video is the easiest way to do it.(4 votes)
1:53Video transcript
- [Voiceover] Let's see
if we can find the limit as x approaches negative infinity of the square root of
4x to the fourth minus x over 2x squared plus three. And like always, pause this video and see if you can figure it out. Well, whenever we're trying to find limits at either positive or negative infinity of rational expressions like this, it's useful to look at what
is the highest degree term in the numerator or in the denominator, or, actually in the numerator
and the denominator, and then divide the
numerator and the denominator by that highest degree,
by x to that degree. Because if we do that,
then we're going to end up with some constants and some other things that will go to zero
as we approach positive or negative infinity,
and we should be able to find this limit. So what I'm talking about,
let's divide the numerator by one over x squared and
let's divide the denominator by one over x squared. Now, you might be saying, "Wait, wait, "I see an x to the fourth here. "That's a higher degree." But remember, it's under the radical here. So if you wanna look at
it at a very high level, you're saying, okay, well x
to the fourth, but it's under, you're gonna take the square
root of this entire expression, so you can really view this
as a second degree term. So the highest degree
is really second degree, so let's divide the numerator and the denominator by x squared. And if we do that, dividing, so this is going to be the same thing as, so this is going to be the limit, the limit as x approaches
negative infinity of, so let me just do a
little bit of a side here. So if I have, if I have one over x squared, all right, let me write it. Let me just, one over x
squared times the square root of 4x to the fourth minus x, like we have in the numerator here. This is equal to, this is the same thing as one over the square
root of x to the fourth times the square root of
4x to the fourth minus x. And so this is equal to the square root of 4x to the fourth minus x over x to the fourth, which is
equal to the square root of, and all I did is I brought
the radical in here. You could view this as the
square root of all this divided by the square root of this, which is equal to, just
using our exponent rules, the square root of 4x
to the fourth minus x over x to the fourth. And then this is the
same thing as four minus, x over x to the fourth is
one over x to the third. So this numerator is going to be, the numerator's going
to be the square root of four minus one, x to the third power. And then the denominator is going to be equal to, well, you divide 2x squared by x squared. You're just going to be left with two. And then three divided
by x squared is gonna be three over x squared. Now, let's think about the limit as we approach negative infinity. As we approach negative infinity, this is going to approach zero. One divided by things that are becoming more and more and more and
more and more negative, their magnitude is getting larger, so this is going to approach zero. This over here is also going to be, this thing is also going
to be approaching zero. We're dividing by larger and
larger and larger values. And so what this is going to result in is the square root of four,
the principal root of four, over two, which is the same thing as two over two, which is equal to one. And we are done.