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AP®︎/College Calculus AB

Course: AP®︎/College Calculus AB>Unit 1

Lesson 15: Connecting limits at infinity and horizontal asymptotes

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.

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• 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.
• 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.
• 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?
• 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.
• At , can someone please explain to me in more detail on how Sal simplified the radical expression in white?
• 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.
• 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.
• 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?
• 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.
• 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?
• 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.
• at how the heck does (4x^4-x)/(x^4) simplify to 4 - (1/x^3) ?? I'm missing something, that for sure.
• 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.
• Hello everyone. I've got a question. In this video we multiply function by 1/x^2 at , 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 "+".