If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

# Limits at infinity of quotients with square roots (odd 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)
Sal finds the limits at positive and negative infinity of x/√(x²+1). Since the leading term is raised to an odd power (1), the limits at positive and negative infinity are different. Created by Sal Khan.

## Want to join the conversation?

• at , why do we take the principal square root of x and not just the square root of x?
• The definition of a function is that an input has one output.

So, if f(x)=sqrt(x), unless we used the principal square root, f(4)= 2 and -2

If this is a function, the input 4 cannot have two outputs!

That is why when using the square root in a function, we use the principal square root.
• Hey, the limit when x-> negative infinite shouldn't be just 1? Since it will be negative infinite divided by negative infinite.
• It's not negative infinity in the denominator because of the absolute value, which comes from taking the principal root of the any number squared.
• At , Sal says that principal root means that we're taking the positive answer.

Is it that we're ignoring the negative, or is it that it doesn't exist (i think it's the first one- but id dont know.) Also if we're ignoring the negative answer, then what will the standard equation look like?

Finally, if we just take the square root, wouldn't it be easier, because then the answer to "limiting x to infinity and negative infinity" will be the same
• Square roots of real numbers are always assumed to be positive. The real number line has a very special property - in analysis it is called being an "ordered field". The ordering means that saying one number is "less than" another, a < b, or "greater than, a > b, makes sense. The complex numbers are not ordered. We can talk about their lengths being > or < but saying "w < z" doesn't make sense. As such (and from convention), we always define the square root to be a positive value, and use the minus/negative sign to explicitly state that the answer is negative. So \sqrt(9) is always 3. If we want -3, we must explicitly state " - \sqrt(9)" to get - 3. As the counterexample, \sqrt(- 5 + 12i) does have meaning, as both 2 + 3i and -2 - 3i = - (2 + 3i) square to get - 5 + 12i. However, since the complex numbers are not ordered, when we talk about the square root or use that symbol, we must recall that both 2 + 3i and its negative are implied. I think Sal is talking about the principal root to remind you that in general, when you square things or take square roots, you need to be aware of the quantity and quality (sign) of your solutions and make sure the notation reflects what you really want to say.

For the second question, standard notation means that the symbol " \sqrt( x^2 + 1) " returns a positive answer. If you want a negative value, you must explicitly state " - \sqrt(x^2 + 1) ". The reason he uses | x | instead of just x is this coupled with order of operations. When you take any x value and square it (and add one), the answer is always positive, which should make sense because you can't take the square root of a negative number. But when we take the square root, it is assumed the resulting value is positive, so we use the absolute value sign to reflect that. If it helps, note that if we allowed both the positive and the negative, the answer would not be x/x, but rather x / ( +/- x) which is no longer a function (one input = two outputs), but instead the implicit graph of two lines.

I hope this helps... underneath your questions are some very deep concepts about the real numbers and our common definition of various symbols.
• How did he conclude that the square root of x was positive? xD
• At , It was the principal square root of x^2 and not just the square root of x.
• How do you calculate the function of a oblique/slant asymptote?
• Not to be disrespectful to Matthew, but I think he made mistakes from ex. (3) (x^3)/(x+2) and on. He consistently left out the constant term of slant asymptotes. For example, he said that the asymptote was y=x^2-2x. But I found it out to be y = x^2 -2x +4.
I did a quick google search on how to find slant asymptote, and I found a simpler method.
http://www.purplemath.com/modules/asymtote3.htm
http://www.purplemath.com/modules/asymnote.htm

Basically you just do a long division, and the asymptote eqn is just the polynomial part of the answer without the remainder, which makes sense bc the remainder is divided by infinity > becomes 0.
• Can someone tell me what an asymptote is?
I don't really get what it is
• The square root of x^2 is +-x. So why did we take the abs value of x? Shouldn't it be +-x
• Also note that we're doing f(x). Functions do not allow ± because we'll end up with 2 y-values for most x-values. Functions give only 0 or 1 y-value per x-value. To avoid any controversy, we simply stick to the "positive square root" (Sal even uses this term in the video).
• How do you determine what the right-end and left-end behavior models of a function will be?
(1 vote)
• How de we know when a function would have a horizontal asymptotic, a vertical asymptotic or both?
• A horizontal asymptote occurs when the limit of the function (the y-value) approaches a constant as x -> infinity (or -infinity). This will happen for rational functions (a polynomial divided by another polynomial) if the degree of the denominator is greater than or equal to the degree of the numerator. It can also occur with special functions like e^x (horizontal asymptote of y = 0 as x -> -infinity).

A vertical asymptote occurs when the function goes to infinity as the x-value approaches some finite number. Usually, this is because there is a fraction where the denominator becomes zero but the numerator does not. For example, f(x) = (x + 2)/(x - 3) has a vertical asymptote of x = 3, since the numerator wants to be 5 and the denominator wants to be 0 at x = 3. Again, there are also some special functions with vertical asymptotes (like ln(x) at x = 0), that you should probably just memorize.
• You showed us
How to solve f(x)=x/Principal square root of x^2+1
But
How to solve f(x)=x/Principal square root of x^3+1
and
How to solve f(x)=x/Principal square root of -x^3+1
(1 vote)
• So we have
f(x) = x / √(x^3 + 1)
= √(x^2) / √(x^3 + 1)
= √(x^2 / (x^3 + 1) )
= √( x^2/x^2 / (x^3/x^2 + 1/x^2) )
= √( 1 / (x + 1/x^2) )
As x goes to infinity, the denominator goes to infinity, so the whole fraction goes to zero and the square root of zero is zero, so f(x) goes to zero.

f(x) = x / √(1 - x^3)
= √( x^2 / (1 - x^3) )
= √( 1 / (1/x^2 - x) )

As x goes to infinity, the denominator goes to negative infinity, but we can't take the square root of a negative number, so f(x) doesn't have a limit as x goes to infinity.