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### Course: AP®︎ Calculus BC (2017 edition) > Unit 1

Lesson 3: Analyzing limits numerically# Approximating limits using tables

In this video, we learn about estimating limit values from tables. The main points are approximating the limit from the left (values less than the target) and the right (values greater than the target). By getting closer to the target value from both sides, we can estimate the limit even if the expression is not defined at the target value.

## Want to join the conversation?

- Just to make sure, what exactly does the number 1.8 represent? It's not the asymptote so is it the Y value as it approaches 3? Does that mean that its going to be open when x=3 ?(21 votes)
- I'm new to this thing too, and because of this this isn't very articulate or eloquently written, but I think it is the value y is approaching as x approaches 3. Well, obviously, that seems quite obvious. But I have to specify this isn't the actual value, since it's actually undefined. In a graph, we represent it with a little circle to show that it isn't actually that number, whether it be undefined or some other number. So even though it's discontinuous, as x is approaching 3, y is approaching (in an intuitive sense) 1.8. It's not the value, but it's where we would graph that part of the line.(22 votes)

- I found a better way to do this to get an exact value. You can simplify the expression to by factoring the numerator and denominator, then canceling the common factor of (x-3) this gives x^2/5 plug in 3 and you get exactly 1.8! I understand the point of the video, but why didn't Sal mention this?(17 votes)
- Excellent point! Yes you are correct that the exact answer of 1.8 can be obtained algebraically. Sal does talk about this technique in a later lesson.(15 votes)

- So what exactly is the purpose of a limit?(7 votes)
- Basically, it allows you to use "infinity" in a sophisticated manner. Ever wondered how on earth we're able to calculate an infinite series in a lifetime?(26 votes)

- so what I did was take out the x squared from the numerator and the 5 from the denom. I then crossed out the (x-3) from the fraction and got x squared over 5. After plugging in 3, I got 1.8. This is something similar to what I saw in my Pre-Calc class, is this a dependable method?(8 votes)
- Yep, that's perfectly correct. You'll eventually see that this is a common method, as the one Sal showed here is just to get an idea for limits. You won't actually take several numbers, find the value of the function at that number, and see what the function approaches for limit problems. What you did is you simplified the expression and substituted the limit. You'll learn this and many more methods soon enough!(10 votes)

- I see these comments are kinda new (23 days highest). How come these calculus videos are being made right now and not before?(4 votes)
- I believe it is KhanAcademy making newer videos to replace the older ones. For example, some of the starting videos in this section are quite old, so I think KA just reshot them to make them better and more current.(11 votes)

- I didn't quite get the idea of asymtote.(2 votes)
- An asymptote is when a line approaches a certain x or y value, but never quite reaches it.

It is hard to explain without having a visual to back me up, but if you have a graphing calculator, then plug in the equation "y = 1/(x^2)". The graph will show that as x approaches both positive infinity and negative infinity, the line approaches, but never touches, y = 0, which shows that an asymptote for this equation is y = 0. Also, as x approches 0, the line never quite reaches x = 0, showing another asymptote which is x = 0.

I would have given a video on the topic, but I couldn't seem to find one, sorry.(5 votes)

- At4:14, I tried doing it on my calculator and it gives me some random answers. It giving me something like 5.99... Is the calculator supposed to be in some specific mode or am I just doing it wrong?(3 votes)
- Without seeing your calculator keystrokes, I don't know for sure what error you are making. Probably the most common error for this type of calculation is to enter the expression incorrectly, so that the calculator performs the operations in the wrong order. Calculator mode is not an issue here for this type of calculation (but degree mode vs. radian mode would have been an issue had the calculation involved a trigonometric function).

Make sure to include parentheses around the expression in the numerator and parentheses around the expression in the denominator, so that the calculator divides these entire expressions.

Also, you should test whether your calculator performs the standard order of operations. For example, if you type 5 + 2 * 3, your calculator performs standard order of operations if it displays 11, but does not do this if it displays 21. It is much preferable to use a calculator that performs standard order of operations.

Example on a TI-89 calculator that performs standard order of operations, for x = 2.999:

the keystrokes (2.999 ^ 3 - 3 * 2.999 ^ 2) / (5 * 2.999 - 15) yield 1.7988002, which is near 1.8.

If you are still getting answers far from the limit value of 1.8 on your calculator, check that you are using the same number of 9's (or same number of 0's) after the decimal point for each instance of x in the function. Also make sure you are including the decimal point for each instance of x in the function.

Have a blessed, wonderful day!(4 votes)

- At2:49, wouldn't any number over three be a negative number? Since evaluating the expression with three gets zero, any number more than three would end in a negative fraction, like -1/-1, which is impossible. Please satisfy my grade 5 mind(1 vote)
- I believe you are mistaken. Sal is saying that dividing by zero is undefined. While having a negative fraction might be unintuitive, it is very real and very possible. If you have two negatives in a fraction, it becomes a positive.(3 votes)

- ist the numerator -54 because it 27-81(2 votes)
- 3 * 3^2 isn't 81, it's 27. So, you will get 0 itself, not -54(3 votes)

- This may be an oversimplification, but it seems to me that determining a limit in calculus if simply finding out when x is this, y is that, i.e., if x = 3, y = 8. Is that right, or is there more to it?(2 votes)
- What you're talking about is just plugging in function values. In a given case, the function value may or may not be the same as the limit.

A limit value is what the function output approaches as the x value approaches a certain number. If this number happens to be the same as the function value, we say the function is continuous at that point. But if the function has a hole or a jump at that point, the limit value will be different.(3 votes)

## Video transcript

- This video we're going to try to get a sense of what the limit
as x approaches three of x to the third minus three x squared over five x minus 15 is. Now when I say get a sense, we're gonna do that by seeing what values for this expression
we get as x gets closer and closer to three. Now one thing that you
might wanna try out is, well what happens to this expression when x is equal to three? Well then it's going to be
three to the third power minus three times three squared, over five times three minus 15. So at x equals three, this
expression's gonna be, let's see in the numerator
you have 27 minus 27, zero. Over 15 minus 15, over zero. So this expression is actually not defined at x equals three. We get this indeterminate form, we get zero over zero. But let's see, even though the function, or even though the
expression is not defined, let's see if we can get a sense
of what the limit might be. And to do that, I'm gonna set up a table. So let me set up a table here. And actually I wanna set up two tables. So this is x and this is x to the third minus three x squared over five x minus 15. And actually, I'm gonna do that again. And I'll tell you why in a second. So this is gonna be x and
this is x to the third minus three x squared over five x minus 15. The reason why I set up two tables, I didn't have to do two tables, I could have done it all in one table, but hopefully this will make it a little bit more intuitive
what I'm trying to do. Is on this left table, I'm
gonna, let's try out x values that get closer and closer
to three from the left. From values that are less than three. So for example, you can
go to two point nine and figure out what the expression equals when x is two point nine. But then we can try to get
even a little bit closer than that, we could go
to two point nine nine. And then we could go
even closer than that. We could go to two point nine nine nine. And so one way to think about it here is as we try to figure out
what this expression equals as we get closer and closer to three, we're trying to approximate
the limit from the left. So limit from the left. Now why do I say the left? Well if you think about
this on a coordinate plane, these are the x values that
are to the left of three, but we're getting closer
and closer and closer. We're moving to the right, but these are the x values that are on the left side of three,
they're less than three. But we also, in order
for the limit to exist, we have to be approaching the
same thing from both sides. From both the left and the right. So we could also try to approximate the limit from the right. And so what values would those be? Well those would be,
those would be x values larger than three. So we could say three point one, but then we might wanna
get a little bit closer, we could go three point zero one. But then we might wanna
get even closer to three. Three point zero zero one. And every time we get
closer and closer to three, we're gonna get a better
approximation for, or we're gonna get a better sense of what we are actually approaching. So let's get a calculator out and do this. And you could keep going,
two nine nine nine nine nine. Three point zero zero zero zero one. Now one key idea here to point out, before I even calculate
what these are going to be, sometimes when people say
the limit from both sides, or the limit from the left
or the limit from the right, they imagine that the limit from the left is negative values and
the limit from the right are positive values. But as you can see here,
the limit from the left are to the left of the x
value that you're trying to find the limit at. So these aren't negative values, these are just approaching the three right over here from
values less than three. This is approaching the three from values larger than three. So now let's fill out this table, and I'm speeding up my
work so that you don't have to sit through me typing everything into a calculator. So based on what we're seeing here, I would make the estimate that this looks like it's approaching one point eight. So is this equal to one point eight? As I said, in the future, we're gonna be able to find this out exactly. But if you're not sure about this you could try an even closer
and closer and closer value.