Main content
Differential Calculus
Course: Differential Calculus > Unit 1
Lesson 15: Limits at infinity- Introduction to limits at infinity
- Functions with same limit at infinity
- Limits at infinity: graphical
- 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
- Limit at infinity of a difference of functions
© 2023 Khan AcademyTerms of usePrivacy PolicyCookie Notice
Limits at infinity of quotients with trig
Sal finds the limit of cosx/(x²-1) at infinity, by putting it between two limits of rational functions, 1/(x²-1) and -1/(x²-1).
Want to join the conversation?
Is there an even mathy way (as suggested in the video by Sal) to prove this? Probably using principle mathematical induction or something?(5 votes)
Hi! This actually does use a "mathy" way to prove, it's called the Squeeze Theorem. Basically, you squeeze an unknown function between two known functions and they trap the unknown function between them. There is a section of videos on it here: https://www.khanacademy.org/math/calculus-home/limits-and-continuity-calc/squeeze-theorem-calc/v/squeeze-sandwich-theorem(23 votes)
I, assume, though he doesn't mention it in the video that we are applying the Pinching Theorem (Squeeze or Sandwich, if you prefer)...(14 votes)
At1:54, why is the inequality -1/(x²-1) <= cosx/(x²-1) <= 1/(x²-1) true for all x?
For example, if x = 0, then -1/((0)²-1) <= cos(0)/((0)²-1) <= 1/((0)²-1) becomes 1 <= -1 <= -1, which is false, right?(6 votes)
You're absolutely right. It seems like a mistake on Sal's part. I think it should have been something like 0 ≤ | cos(x)/(x^2 -1) | ≤ | 1/(x^2 -1) |. Good catch.(3 votes)
- wait, so irrespective of the sign of the numerator, as long as the value of the denominator becomes increasingly negative or increasingly positive, the limit as x --> infinity & x--> -infinity will equal to zero?(3 votes)
- Yes, you are correct. But to be clear, as long as the denominator becomes sufficiently LARGE as compared to a relatively small numerator (whether positive or negative), the limit as x->infinity will be 0.
Remember, a tiny numerator (negative or positive) divided by a HUGE denominator (negative or positive) will be very close to zero. That's why the limit as the number goes to infinity is zero.(4 votes)
- how does it become a negative?...when its to the power of an odd ?(1 vote)
Hi! Are you asking about why the cos(x) can be negative? This has to do with what cosine is. In short, cosine is a trig function that is constantly moving between -1 and 1.
Here is a picture of cos(x) along with sin(x):
https://upload.wikimedia.org/wikipedia/commons/thumb/7/71/Sine_cosine_one_period.svg/2000px-Sine_cosine_one_period.svg.png
We want to look at the blue function. Without getting too in depth, we can see that its y value is always between -1 and 1, this is why Sal makes it negative. Feel free to check out the trig section on KA to learn more!(5 votes)
- there is a mistake: you can't say that if -1<cosx<1 then -1/x*2-1<cosx/x*2-1<1/x*2-1; because if x*2-1 is negative than you would have to invert the diseqution(3 votes)
He's talking about their "limiting value" when x approaches infinity :)(1 vote)
At x equals +1 or -1, the function goes undefined. Shouldn't we consider that point as we approach Infinity ?(2 votes)
Your point is correct. However, when we talk about taking the limit "as x approaches infinity," we should really be thinking of "the value of the function AT x = infinity." Also, limits are only concerned with the point we end at (infinity in this case), not where we "start approaching from." Anyway, when we think of limits "AT" infinity, we plug in x = infinity into the expression, remembering that p*x^a + q*x^b =
(1) p*x^a if a > b,
(2) q*x^b if b > a, or
(3) (p + q)*x^a if a = b.
The above conditions hold true at x = infinity and allow us to reduce the problem to a limit that is smaller in magnitude than 1/x^2, which abruptly flattens to 0 as x becomes large.(3 votes)
it is like squize theorem , right?(1 vote)- Is this Squeeze theorem ?(1 vote)
- I have a question from parametric equations for curves. The question is "3 over theta equals infinity".
Shouldn't it be zero?(1 vote)
1:54
Video transcript
- [Voiceover] So, let's
see if we can figure out what the limit as X approaches infinity of cosine of X over X
squared minus one is. And like always, pause this video and see if you can work
it out on your own. Well, there's a couple
of ways to tackle this. You could just reason
through this and say, "Well, look this numerator,
right over here, cosine of X, "that's just going to oscillate between "negative one and one." Cosine of X is going to
be greater than or equal to negative one, or negative
at one is less than or equal to cosine of X which is
less than or equal to one. So, this numerator just oscillates between negative one and one as X changes, as X increases in this case. While the denominator
here, we have an X squared, so as we get larger and larger X values, this is just going to become
very, very, very large. So, we're going to have something bounded between negative one and one divided by very, very infinitely large numbers. And so, if you take a, you
could say, bounded numerator and you divide that
infinitely large denominator, well, that's going to approach zero. So, that's one way you
could think about it. Another way is to make this same argument, but to do it in a little
bit more of a mathy way. Because cosine is bounded in this way, we can say that cosine of X over X squared minus one is less than or equal to. Well, the most that this
numerator can ever be is one, so it's going to be less
than or equal to one over X squared minus one. And it's going to be a
greater than or equal to, it's going to be greater than or equal to, well, the least that this
numerator can ever be is going to be negative one. So, negative one over X squared minus one. And once again, I'm just saying, look, cosine of X, at most, can be one and at least is going to be negative one. So, this is going to be true for all X. And so, we can say that also the limit, the limit as X approaches infinity of this is going to be true for all X. So, limit as X approaches infinity. Limit as X approaches infinity. Now, this here, you could
just make the argument, look the top is constant. The bottom just becomes infinitely large so that this is going to approach zero. So, this is going to be
zero is less than or equal to the limit as X approaches infinity of cosine X over X squared minus one which is less than or equal to. Well, this is also going to go to zero. You have a constant numerator,
an unbounded denominator. This denominator's
going to go to infinity, and so, this is going to be zero as well. So, if our limit is
going to be between zero. If zero is less than
or equal to our limit, is less than or equal to zero, well then, this right over here has to be equal to zero.