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# Limits at infinity of quotients (Part 1)

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 (4x⁵-3x²+3)/(6x⁵-100x²-10). Created by Sal Khan.

## Want to join the conversation?

• In the dominating term of the polynomial where the x is raised to the fifth power, and you let x be negative, you end up with a negative number, as it does for all odd powers of x. So, it seems to me that you end up in the limit having a very large negative number divided by a very large negative number, which will be a positive number determined by the ratio of coefficients. What is wrong with my thinking?.
• There is nothing wrong with your thinking. The lim x->-∞ = +2/3
Here is a more mathematical way of thinking about these limits. If you multiply each term by 1/x^n (where n is the highest degree term in the function) the limit can be evaluated. For example,
(4x^5-3x^2+3)/(6x^5-100x^2-10) * (1/x^5) / (1/x^5)
(4+-3x^-3+3x^-5)/(6-100x^-3-10x^-5)
The above limit can be evaluated. All of the terms with x to a negative power will approach 0 as x goes to +∞ and -∞. The limit simplifies to
(4+0+0)/(6+0+0) = 2/3
• Is it safe to assume that the limit as x -> ∞ of any standard polynmial in the case of a polynomial fraction (so to speak) where the highest degree power of the numerator is equal to that of the denominator; it is the ratio of the coefficient of the highest degree powers in both the numerator & denominator? And the case where the numerator's highest degree is less than the denominator's; the limit is 0 (i.e the funtion is approaching 0 as x -> ∞) because the denominator overpowers the numerator? But then if the highest degree power in the denominator is less than the highest degree power in the numerator, the denominator will obviously be overpowered by the numerator. Going further into that, if you employ the neat little trick of dividing all the terms in the function by the highest degree power of x, the denominator would be seen to approach 0 as we get to infinity and we know that we cannot have a denominator value of 0. Is this when some sort of algebraic simplification is required so as to determine the limit as the function approaches infinity? Or is there a massive flaw in my reasoning? Thank you :)
• First, remember that with limits we are not evaluating what the function is at the limiting value, but rather what the function approaches as we get infinitesimally close to the limiting value. This distinction is important because having a denominator approaching 0 is not the same thing as having a denominator actually at 0. Thus, it is possible to evaluate a limit when the denominator approaches 0, because we are not actually dividing by 0, but by something exceedingly close to 0.

Anyway, here are the various scenarios for the limit as x→∞ for rational functions (that is what you call a polynomial divided by a polynomial).
Look at only the highest power terms in both the numerator and denominator, ignore all other terms. Note their coefficients. Let us call the coefficient in the numerator n and that in the denominator we'll call d.
Scenario 1: If the numerator has the higher power while n and d have the same sign, then the limit is +∞
Scenario 2: If the numerator has the higher power while n and d have different signs, then the limit is -∞
Scenario 3: If the denominator has the higher power, then the limit is 0.
Scenario 4: If the numerator and denominator have the same highest power, then the limit is a/b.

Note: these simple ways of solving limits only work for rational functions. If you have more complicated functions, you may need to use more sophisticated means of evaluating the limit such as l'Hopital's Rule.
• What does e stand for? In several of the problems in "Limits at infinity where x is unbounded, I run across the letter e. I also find ln. Does anybody know what these stand for? Do they have specific values? And can you refer me to a KA video that will tell me the formal definitions and/or proofs?
• Yes, I can tell you.
e is a constant that (like π) comes up frequently in all sorts of math and science. It is known as Euler's number. It is an irrational and transcendental number. It is officially defined as:
`e = lim h→0 (1+h)^(1/h)`
The first few digits of e are 2.71828...

`ln` is the natural logarithm, that is it is log base e. `ln x` is the inverse function of e^(x). In other words:
`ln(e^x) = x` and `e^(ln x) = x`
The natural logarithm may be defined as:
`ln x = lim h→0 [x^(h) - 1]/h`

You should have covered the basics of e and logarithms in a previous course such as Algebra II or Trigonometry/Precalculus. The ability to work with exponential functions and logarithms is prerequisite to studying calculus, so you may need to review that material.
• How do we know whether f(x) would have a vertical or horizontal asymptote?
• You have a vertical asymptote at x=c if:
lim x→c⁻ f(x) = + ∞ or −∞
or lim x→c⁺ f(x) = + ∞ or −∞
(only has to be one of these conditions, but can be both).

You have a horizontal asymptote, f(x) = n if there exists a finite real number, n, such that
lim x→−∞ f(x) = n or lim x→ +∞ f(x) = n
(just has to be one of these conditions, but can be both).

Most of the time, you have a vertical asymptote due to division by 0
or by tan (½kπ) where k is an odd integer.
• how do you find the limit as x>infinity of (4^x+7^x)^(1/x)
• This is easily solved by rewriting the expression in a useful form. We may factor out a `7ᵡ` from the "inside" as follows:

`(4ᵡ + 7ᵡ)^(1/x) = [7ᵡ(1 + (4/7)ᵡ)]^(1/x) = 7 · (1 + (4/7)ᵡ)^(1/x).`

The second factor on the right-hand side clearly goes to `1` as `x → +∞`. Therefore, the entire expression goes to `7` as `x → +∞`.
• Isn't there another vertical asymptote at between x=3 and x=4?
• There is a vertical asymptote close to `x=2.5672`. But vertical asymptotes are not considered when finding limits at infinity.
• can you solve it for lim x-> -∞ of ln(x-1)/x
• We may assume `x < 1` (why?). For such `x`, we have `x - 1 < 0`, so

`log(x - 1)`
`= log(|x - 1|) + iArg(x - 1)`
`= log(1 - x) + iπ.`

Hence `log(x - 1) = log(1 - x) + iπ` for `x < 1`. Since `log(1 - x)/x → 0` as `x → -∞`, it follows that

`log(x - 1)/x = log(1 - x)/x + iπ/x → 0` as `x → -∞`.

(To see why `lim (x → -∞) log(1 - x)/x = 0`, apply L'Hôpital's rule.)

We conclude that
`lim (x → -∞) log(x - 1)/x = 0.`
• what is "This thing" at he says " as X gets very very large this thing is going to approximate" is "this thing" just a squiggly 'equals' sign?
• The '≈' sign means 'approximately equal to'. I assume you know that π is an irrational number with infinitely many decimals, and that it is approximately equal to 3.14159265… etc. Then we can write π ≈ 3.14159265. It is not equal to that value, but the approximation works for most practical situations.
• So Sal basically what you are telling us that if we find horizontal asymptote of a rational function then we can find its limit when x goes to negative or positive infinity ??
• Exactly, unless there is no horizontal asymptote, in which case it is the oblique asymptote