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Limits at infinity of quotients (Part 1)

Sal finds the limits at positive and negative infinity of (4x⁵-3x²+3)/(6x⁵-100x²-10). Created by Sal Khan.

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  • blobby green style avatar for user Bill Cowhig
    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?.
    (41 votes)
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    • spunky sam blue style avatar for user Ethan Dlugie
      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
      (86 votes)
  • blobby green style avatar for user John Uduma Junior
    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 :)
    (10 votes)
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    • piceratops ultimate style avatar for user Just Keith
      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.
      (23 votes)
  • primosaur ultimate style avatar for user GodIsREAL
    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?
    (6 votes)
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    • piceratops ultimate style avatar for user Just Keith
      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.
      (12 votes)
  • blobby green style avatar for user Laura Kim
    How do we know whether f(x) would have a vertical or horizontal asymptote?
    (5 votes)
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    • piceratops ultimate style avatar for user Just Keith
      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.
      (5 votes)
  • leaf green style avatar for user joe
    how do you find the limit as x>infinity of (4^x+7^x)^(1/x)
    (4 votes)
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    • leaf grey style avatar for user Qeeko
      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 → +∞.
      (5 votes)
  • blobby green style avatar for user sc28
    Numerator = Denominator, then the limit is simply the coefficients. If the numerator > denominator, then the limit is at infinity. Lastly, if the numerator is less than than the denominator, then the limit is 0. Remember we are talking about degrees here. So compare the numerator and denominator in terms of degrees. Hope that helps! @Savannah
    (5 votes)
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  • aqualine tree style avatar for user Rupak
    Isn't there another vertical asymptote at between x=3 and x=4?
    (3 votes)
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  • leaf green style avatar for user kanyutha
    can you solve it for lim x-> -∞ of ln(x-1)/x
    (2 votes)
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    • leaf grey style avatar for user Qeeko
      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.
      (5 votes)
  • blobby green style avatar for user bblough
    Why would L'Hopital's Rule not work for these problems? I tried to apply it to some of the practice problems in this section and had no success.
    (1 vote)
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    • old spice man green style avatar for user Elijah Daniels
      Remember that in order to apply L'Hopital's rule, you have to evaluate the limit of the function as x approaches infinity first and get an indeterminate form (0/0, infinity/infinity, and others). If you evaluate the limit and get an indeterminate form, you can apply L'Hopital's rule. If you do not get an indeterminate form, you cannot apply L'Hopital's rule.
      (5 votes)
  • aqualine ultimate style avatar for user Ridwan Ahmed
    How do you find out whether the function is approaching the limiting value from the positive or negative direction? For example, in the problem posed above, how do you determine that the function approaches 2/3 from the bottom when x -> -infinity and from the top when x -> +infinity?
    (3 votes)
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    • female robot grace style avatar for user loumast17
      For rational functions like this you would want to find the greatest vertical asymptote and then calculate if the function at a value greater then that is going to be above or below the horizontal asympotoe. Alternatively you could just use a really really big number to be sure. If the output for that is below the horizontal asympote it will approach from below, and above means it approaches from above.
      (1 vote)

Video transcript

So we have f of x equaling 4x to the fifth minus 3x squared plus 3, all of that over 6x to the fifth minus 100x squared minus 10. Now, what I want to think about is-- what is the limit of f of x, as x approaches infinity? And there are several ways that you could do this. You could actually try to plug in larger and larger numbers for x and see if it seems to be approaching some value. Or you could reason through this. And when I talk about reasoning through this, it's to think about the behavior of this numerator and denominator as x gets very, very, very large. And when I'm talking about that, what I'm saying is, as x gets very, very large-- let's just focus on the numerator. As x gets very, very large, this term right over here in the numerator-- 4x to the fifth-- is going to become a much, much more significant than any of these other things. Something squaring gets large. But something being raised to the fifth power gets raised that much, much faster. Similarly, in the denominator, this term right over here-,, the highest degree term-- 6x to the fifth-- is going to grow much, much, much faster than any of these other terms. Even though this has 100 as a coefficient or a negative 100 as a coefficient, when you take something to the fifth power, it's going to grow so much faster than x squared. So as x gets very, very, very large, this thing is going to approximate 4x to the fifth over 6x to the fifth for a very large, large x Or we could say as x approaches infinity. Now, what could this be simplified to? Well, you have x to the fifth divided by x to the fifth. These are going to grow together. So these you can think of them as canceling out. And so you are left with 2/3. So what you could say is-- the limit of f of x, as x approaches infinity, as x gets larger and larger and larger, all of these other terms aren't going to matter that much. And so it's going to approach 2/3. Now, let's look at the graph and see if that actually makes sense. What we're actually saying is that we have a horizontal asymptote at y is equal to 2/3. So lets look at the graph. So right here is the graph. Got it from Wolfram Alpha. And we see, indeed, as x gets larger and larger and larger, f of x seems to be approaching this value that looks right at around 2/3. So it looks like we have a horizontal asymptote right over here. Let me draw that a little bit neater. We have a horizontal asymptote right at 2/3. So let me draw it as neatly as I can. So this right over here is y is equal to 2/3. The limit as x gets really, really large, as it approaches infinity, y is getting closer and closer and closer to 2/3. And when we just look at the graph here, it seems like the same thing is happening from the bottom direction, when x approaches negative infinity. So we could say the limit of f of x, as x approaches negative infinity, that also looks like it's 2/3. And we can use the exact same logic. When x becomes a very, very, very negative number, as it becomes further and further to the left on the number line, the only terms that are going to matter are going to be the 4x to the fifth and the 6x to the fifth. So this is true for very large x's. It's also true for very negative x's. So we could also say, as x approaches negative infinity, this is also true. And then, the x to the fifth over the x to the fifth is going to cancel out. These are the dominant terms. And we're going to get it equaling 2/3. And once again, you see that in the graph here. We have a horizontal asymptote at y is equal to 2/3. We take the limit of f of x as x approaches infinity, we get 2/3. And the limit of f of x as x approaches negative infinity is 2/3. So in general, whenever you do this, you just have to think about what terms are going to dominate the rest? And focus on those.