AP®︎/College Calculus AB
Sal analyzes the behavior of f(x)=x/[1-cos(x-2)] around its asymptote at x=2.
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- what happens when you're given f(x)= 3e^-x + 2 ? how do you determine the limit at infinity ?(4 votes)
- 3e^(-x)=3/(e^x). So as x gets very large, e^x gets incredibly large, and so 3/(e^x) goes to 0.
So you're just left with 2.(20 votes)
- Does that mean limit exist at x= 2?(4 votes)
- Is it possible to think about this cosine function using a secant graph?(4 votes)
- Picture the two graphs in your head: https://www.khanacademy.org/computer-programming/cos-sec-comparison/6530932543127552
There a couple things you can notice right of the bat:
- The graph of cos(x) is continuous, while sec(x) is not.
- The slope of sec(x) is always the opposite sign of the slope of cos(x) (m ≠ 0).
- The y coordinates of cos(x) and sec(x) are never the same for the same x coordinate except when there is a relative max/min.
- Because the graphs of cos(x) and sec(x) always touch each other at their relative maximums and minimums, if you are given the graph of one you can visualize the graph of the other.
So while you can visualize the graph of cos(x) from looking at the graph of sec(x) (and vice versa), you cannot reliably do any calculations with sec(x) and expect to get the same result for cos(x).
I hope this helps!(7 votes)
- How would you do that sort of problem without a calculator? Is there any algebra you could use?(6 votes)
- When I did this problem, I did not use any algebra per se to get my final answer, But I did not resort to using a calculator either.. I noted that as x approaches two from both directions(left and right), cos(x-2) must approach 1 from values less than 1. Hence, 1-cos(x-2) must approach 0 from values greater than 0. So as the numerator is approaching 2, the denominator is approaching 0 from the right. This means the expression must approach infinity because values close to 2 are being divided by values that are infinitesimally close to and to the right of 0.(2 votes)
- ¿How can the cosine be always positive?(4 votes)
- It isn't, but when x is close to 2 the value of
cos(x-2)will be near
cos(0)i.e. near to 1.
Most importantly, unless
x=2the value of
cos(x-2)will always be less than 1. This means that the denominator
1 - cos(x-2)will always be positive for values of x that are close to, but not equal to 2.(5 votes)
- Does 0.99999999999999999999999999999999=1?(0 votes)
- No, it doesn't, unless you mean 0.9 repeating. 0.99999999999999999999999999999999 or whatever you said only approaches 1.(3 votes)
- Shouldn't the limit of the function, that is in the video, be either be undefined, unbounded, or nonexistent, instead of being equal to infinity? Also, isn't infinity not a number, but a concept. It is just so confusing. Can anyone help me understand?(2 votes)
- You're correct that the limit doesn't exist if the function goes to infinity. But if the function goes to infinity or negative infinity, we often like to write the limit as such because it provides more information. It's still implied that the limit doesn't exist, we're just giving info about the functions end behavior as well.(3 votes)
- There is another way.
1-cos(x-2) is always positive because cos never gets bigger than 1. So the denominator is always positive.
The numerator isn't. When x>0 the numerator is positive and so is the function, and when x<0 the numerator is negative and so is the function. And we know that the function has unbounded limits (from the left and right) when x approaches 2. We can immediately conclude that those limits are both positive, because 2 is bigger than 0.(1 vote)
- How do you evaluate for a function like this:
f(x) = -x/ln^2(x-1)
I am trying to evaluate for when x approaches 2 from the left and right. I used the hints in the practice problems and did not understand how they obtained their values. Could someone please show me how to evaluate a function like this?
Thank You!(1 vote)
- Hi there
Use the idea that that ln(1) =0, and that for x>1, ln(x) is positive.
As x approaches 1 from the right, the values of ln(x) will become very small positive numbers.
So now, the numerator will have a value close to -1, while the denominator has a small positive value that you will square. The limit will be negative infinity.
Approach 1 from the left, and now ln(x) will approach a very small negative number - but when you square it, you will get a positive value - the left hand limit will equal the right hand limit. Now, while the limit "technically" doesn't exist at x = 1 because of the vertical asymptote, we do write that the limit is negative infinity... that at least tells us how the limit fails to exist and that the graph of the function has a vertical asymptote at x = 1.(1 vote)
- [Voiceover] So we're told that f of x is equal to x over one minus cosine of x minus two, and they ask us to select the correct description of the one-sided limits of f at x equals two. And we see that right at x equals two, if we try to evaluate f of two, we get two over one minus cosine of two minus two, which is the same thing as cosine of zero, and cosine of zero is just one, and so one minus one is zero, and so the function is not defined at x equals two, and that's why it might be interesting to find the limit as x approaches two, and especially the one-sided limit. And the one-sided limit. Well, we'll obviously leave it at that so let's try to approach this. So there's actually a couple of ways you could do it. There is one way you could do this without a calculator by just inspecting what's going on here, and thinking about the properties of the cosine function, and if that inspires you, pause the video and work it out, and I will do that at the end of this video. The other way, if you have a calculator is to do it with a little bit of a table like we've done in other example problems. So if we think about x approaching two from the positive direction, well then. We can make a little table here where you have x, and then you have f of x. And so for approaching two from values greater than two, you could have 2.1, 2.01. Now the reason why I said calculators, these aren't trivial to evaluate because this would be what, 2.1 over 1 minus cosine of, 2.1 minus two is 0.1. I do not know what cosine of 0.1 is without a calculator. I do know that cosine of zero is one so this is very, very close to one without getting to one, and it's going to be less than one. Cosine is never going to be greater than one. The cosine function is bounded between negative one is less than cosine of x. I'll just write the x, then I don't need the parenthesis. Which is less than one. The cosine function just oscillates between these two values so this, this thing is gonna be approaching one but it's going to be less than one. It definitely cannot be greater than one, and that's actually a good hint for how you can just explore the structure here, and then you could say, "All right, 2.01. "Well, that's going to be 2.01 "over one minus "cosine of 0.01." And this is going to even closer to one without being one but it's going to be less than one. No matter what, cosine of anything is going to be between negative one and one, and it could even be including those things but as we approach two, this thing is going to approach one, I guess you could say approach one from below. And so you can start to make some intuitions here. If it's approaching from below, this thing over here, this whole expression is going to be positive, and as we approach x equals two. Well, the numerator is positive. It's approaching two. The denominator is positive so this whole thing has to be approaching a positive value or it could become unbounded in the positive direction as we'll see, this is unbounded because this thing is even closer to one than this thing, and you would see that if you have a calculator but needless to say, this is going to be unbounded in the positive direction so we're going to be going towards positive infinity so these two choices have that, and we can make the exact same argument as we approach x in the negative or from below, as we approach two from below, I should say. So that's x, and that is f of x, and once again, I don't have a calculator in front of me. You could evaluate these things at a calculator, and become very clear that these are positive, and as we get closer to, they become even larger and larger positive values, and the same thing would happen if you did 1.9, and if you did 1.99 because here, you'd be 1.9 over one minus cosine. Now here, you'd have 1.9 minus two so this would be negative 0.1. Let me scroll over a little bit. The second one would be 1.99 over one minus cosine of negative 0.01. And cosine of negative 0.1 is the same thing as cosine of 0.1. Cosine of negative 0.01 is the same thing as cosine of 0.01. So these two things, this is going to be equal to that. That is going to be equal to that. And once again, we're gonna be approaching positive infinity so the only choice where all of that is true is this first one. Whether we approach two from the right-hand side or the left-hand side, we're approaching positive infinity but the other way you could have deduced that is say, "Okay, as we approach two, the numerator is going to be positive 'cause two is positive, and then over here, as we approach two, cosine of anything can never be greater than one. It's going to approach one but be less than one so if this is less than as x approaches two. It becomes one when x is equal to two. Well then, this right over here, one minus something less than one is going to be positive so you have a positive divided by a positive so you're definitely going to get positive values as you approach two. And we know, and they've already told us that these are going to be unbounded based on the choices so you would also pick that but you should also feel good about it, that the closer that we get to two, the closer that this value right over here gets to zero. And the closer that this value gets to zero, the closer we get to one. The closer we get to one, the smaller the denominator gets. And then you divide by smaller and smaller denominators, you're going to become unbounded towards infinity, which is exactly what we see in that first choice.