AP®︎ Calculus BC (2017 edition)
- 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
Sal analyzes the limits at infinity of three different rational functions. He finds there are three general cases of how the limits behave. Created by Sal Khan.
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- I understand how he gets to the limits and all, but can someone give me a 100% accurate definition of what exactly a "asymptote" is? what is an asymptote? Is it just the graph of the limit? please help me !(43 votes)
- This is actually not quite accurate.
Technically an asymptote to a curve is a line such that the distance between the curve and the line approaches zero as they both tend toward infinity.
A function can never cross a vertical asymptote, but it may cross a horizontal asymptote an infinite number of times (thus reaching it an infinite number of times). For example observe the limit of sin(x)/x as x approaches infinity.(25 votes)
- Are + and - infinity like approaching from the positive and negative sides? So limits approaching infinity dont exist, because + and - infinity yield different values?(9 votes)
- Not exactly,
positive and negative infinity represent the opposite "ends" of the number line. And here, "ends" is in quotation marks because the number line NEVER actually ends, it goes on forever in both directions. Basically positive infinity means to keep going towards bigger and bigger positive numbers. Think of the biggest positive number you can think of, and then go even bigger than that… and keep doing that… FOREVER! That's positive infinity.
For negative infinity, think of the most negative number you can think of, and then think of an even more negative number, and keep doing that, FOREVER.
So you see, if a limit approaches positive infinity from one side, and negative infinity from the other side… it doesn't approach the same thing from both sides. THIS is why the limit doesn't exist. It would be the same as saying that a limit that approaches 3 from the positive side and 2 from the negative side also doesn't exist. In order for a limit to exist it must approach the same thing from both sides.(28 votes)
- Where exactly can we use the concept of limits in our life?
In what kind of situations in our life do we need to find limits?
Please give me a real life example where in we have to find the limit.
Thanks in advance!(8 votes)
- Limits are not that helpful in everyday life. However, as you study Calculus, you'll see that every single concept in Calculus in based upon limits. Derivatives are defined using limits, we can find the area under a curve, volume, surface area, arc length, radius of curvature using integrals that are all based upon limits. Without limits, there is no Calculus.(5 votes)
- In the solutions manual of my Calculus textbook, it gets the answer using a slightly different method. It divides like every term in the numerator and the denominator by the highest degree i guess and does all these weird calculations and then gets the answer. I understand this method much easier though (getting the highest power of both the numerator and denominator and then applying the x-> component) I guess my question is if I do this method that you are teaching on a test, is it still valid and legitimate? Of will i have to provide more work to justify my answer? Sorry for the incredibly long comment. Thanks.(10 votes)
- Sal's method is more of a "rationalize it" (no pun intended) approach. What your textbook says is the method I prefer, mainly because it gives you a clear, algebraic way to evaluate the limit.
As for which method to use on a test, that is something you should ask your teacher.(9 votes)
- So then would it be safe to assume that any time the numerator is growing faster than the denominator it will equal infinity, and any time the denominator is growing faster than the numerator it will equal 0? Or is there an exception?(12 votes)
Notice that we are dealing with x nearly equal to Infinity.
Which is very very big. Thus we can safely assume.
But in case of small numbers like 2, 56, 345.... We cant assume so.
You can try it out yourself.(4 votes)
- At2:00how can he simplify that to 1/2x when there are two different powers?(7 votes)
- He cancelled out common factors
3x^3/6x^4 can be factored to
You can cancel out the three and three of the x terms
( 3*x*x*x )/(2* 3*x*x*x *x)
This leaves 1/(2*x)(10 votes)
- at3:20, is the simplification of (4/250)x to (2/125)x valid ?(4 votes)
- Yes, but it is unnecessary. With limits, since you often have them diverge toward +∞ or −∞ or else tend toward 0, you can save yourself unnecessary work by not simplifying any constants until you know you don't have an infinity or zero situation. When tending toward 0, your constant is irrelevant and there is no need to simplify. When tending toward ∞, you need only determine the sign of your constant, to determine whether you're tending toward + or − infinity.
It is only when you're tending toward a non-zero but finite number that you need to simplify your constant.
Note: The above applies to real numbers. If imaginary numbers get involved, the considerations can be more complicated.(8 votes)
- Lim(x--->infinity) (1+1/n)^n =e
How to do solve these very strange infinity limits or either to prove them they equal to e?(4 votes)
x (1 + 1/x)^x (1 + 1/x)^(1/x)
1 2.000000000 2.00000000000000
10 2.593742460 1.00957658300000
100 2.704813829 1.00009950825915
1000 2.716923932 1.00000099950083
10000 2.718145927 1.00000000999950
100000 2.718268237 1.00000000010000
1000000 2.718280469 1.00000000000100
10000000 2.718281694 1.00000000000001(7 votes)
- Will a similar method be used for an expression in which there is no denominator like for limit x tending to infinity √(n-1) - √n
Here will it be correct to say that this is the same as √n - √n (As the 1 will be negligible when n tends to infinity) and thus the limit equals 0 ?(4 votes)
- Yep. That general pattern of thinking is correct; actually, though, the example you've mentioned is a bit interesting in the sense that we can't always define a limit that takes the form ∞ - ∞. In this case, because the two terms are of the same degree, the limit is equal to 0 (and a quick glance at the graph of
y = sqrt(x-1) - sqrt(x)confirms that as x approaches infinity, y approaches 0). As you said, it resembles
y = sqrt(x) - sqrt(x) = 0in the limit.
Other limits of a similar nature may not always behave the same way. Take the limit of
x^3 - x^2as x approaches infinity, and we get infinity rather than 0 because the terms are of a different degree (which seems fairly clear just by looking at the function). Sometimes the examples are less clear-cut, so it's worth exercising some caution with limits of the form ∞ - ∞.
I hope you find that helpful; you're definitely on the right track.(6 votes)
- I understand how he got the second example and I ended up with the same answer myself. My method was to take a x^4 term in both the numerator and denominator ( 0x^4 /6x^4). That simply leaves us with zero and I wanted to know whether this is a perfectly viable method? I do understand the issue it would cause if taking the reciprocal (divide by zero).(5 votes)
- This is a perfectly viable method, and is often taught as a shortcut to the process of taking limits at infinity, taking the quotient of the terms with highest power in the numerator/denominator. In the case of taking the reciprocal, the limit would go to infinity (which will be covered in a later topic).(3 votes)
Let's do a few more examples of finding the limit of functions as x approaches infinity or negative infinity. So here I have this crazy function. 9x to the seventh minus 17x to the sixth, plus 15 square roots of x. All of that over 3x to the seventh plus 1,000x to the fifth, minus log base 2 of x. So what's going to happen as x approaches infinity? And the key here, like we've seen in other examples, is just to realize which terms will dominate. So for example, in the numerator, out of these three terms, the 9x to the seventh is going to grow much faster than any of these other terms. So this is the dominating term in the numerator. And in the denominator, 3x to the seventh is going to grow much faster than an x to the fifth term, and definitely much faster than a log base 2 term. So at infinity, as we get closer and closer to infinity, this function is going to be roughly equal to 9x to the seventh over 3x to the seventh. And so we can say, especially since, as we get larger and larger as we get closer and closer to infinity, these two things are going to get closer and closer each other. We could say this limit is going to be the same thing as this limit. Which is going to be equal to the limit as x approaches infinity. Well, we can just cancel out the x to the seventh. So it's going to be 9/3, or just 3. Which is just going to be 3. So that is our limit, as x approaches infinity, in all of this craziness. Now let's do the same with this function over here. Once again, crazy function. We're going to negative infinity. But the same principles apply. Which terms dominate as the absolute value of x get larger and larger and larger? As x gets larger in magnitude. Well, in the numerator, it's the 3x to the third term. In the denominator it's the 6x to the fourth term. So this is going to be the same thing as the limit of 3x to the third over 6x to the fourth, as x approaches negative infinity. And if we simplified this, this is going to be equal to the limit as x approaches negative infinity of 1 over 2x. And what's this going to be? Well, if the denominator, even though it's becoming a larger and larger and larger negative number, it becomes 1 over a very, very large negative number. Which is going to get us pretty darn close to 0. Just as 1 over x, as x approaches negative infinity, gets us close to 0. So this right over here, the horizontal asymptote in this case, is y is equal to 0. And I encourage you to graph it, or try it out with numbers to verify that for yourself. The key realization here is to simplify the problem by just thinking about which terms are going to dominate the rest. Now let's think about this one. What is the limit of this crazy function as x approaches infinity? Well, once again, what are the dominating terms? In the numerator, it's 4x to the fourth, and in the denominator it's 250x to the third. These are the highest degree terms. So this is going to be the same thing as the limit, as x approaches infinity, of 4x to the fourth over 250x to the third. Which is going to be the same thing as the limit of-- let's see, 4, well I could just-- this is going to be the same thing as-- well we could divide two hundred and, well, I'll just leave it like this. It's going to be the limit of 4 over 250. x to the fourth divided by x to the third is just x. Times x, as x approaches infinity. Or we could even say this is going to be 4/250 times the limit, as x approaches infinity of x. Now what's this? What's the limit of x as x approaches infinity? Well, it's just going to keep growing forever. So this is just going to be, this right over here is just going to be infinity. Infinity times some number right over here is going to be infinity. So the limit as x approaches infinity of all of this, it's actually unbounded. It's infinity. And a kind of obvious way of seeing that, right, from the get go, is to realize that the numerator has a fourth degree term. While the highest degree term in the denominator is only a third degree term. So the numerator is going to grow far faster than the denominator. So if the numerator is growing far faster than the denominator, you're going to approach infinity in this case. If the numerator is growing slower than the denominator, if the denominator is growing far faster than the numerator, like this case, you are then approaching 0. So hopefully you find that a little bit useful.