In the last question, how does x→7 lim g(x) exist? It has two locations right?

The limit exists because the same y-value is approached from both sides. It does not have two locations because the open circle is a just gap in the graph. The closed circle is the actual y-value for when x=7.

In problem 5 why can one of the answers be x→6? but not x→3?

I assume you are talking about the last example. As you approach x=6 from the left you move closer to 3; AND as you approach x=6 from the right, you also moce closer to 3. As you move closer to x=3 form the left you move closer to 3, BUT when you move closer to x=3 from the right, you move closer to 6. They must be moving to the same value of y from both sides if you are not going to specify the side in th limit notation.

learning this calculus concept here was actually super fun. although i know this will take many hours, the hours will be exhilarating! i'm going to try and conquer calculus in 2 weeks. (today is june 24th, i have until july 10th)

How is calc going so far for ya'll

For the last question, how is there a limit for x-->6? There is a point there. I thought there could only be limits if there were open dots.

As Sal explained both in the video Limits intro, and in the text, the beauty of limits, and one property of limits, is that they do not explain the actual point of the graph, but the behavior leading up to that point. Whether there is a point at f(x)= 6 or a hole, that would not change that there still is a limit, unless a jump occurs between the two leading lines.

When we take a limit, we approach a specific x value from both sides. But there is _an infinite number of numbers between any two numbers_. So when we are approaching x, *can* *we* *ever* *really* *get* *there*? And if we can not, then does the calculus give us an answer with really really really really really small flaw that does not create problem in any calculation? Or is it 100% flawless?

This is the issue that a lot of people had in the development of the calculus. The solution was quite clever, the idea is that you can get "arbitrarily close" that is given any epsilon positive there's a delta such that |x-a| < delta implies |f(x) - f(a)| < epsilon. If this is true FOR EVERY epsilon, then we say the limit exists and the function is continuous. Is it 100% flawless? Not really, but the thing is it's useful. Using these ideas we were able to build up a foundation for calculus which eventually lead to the physics that put man on the moon. The point is it gets results.

the plot has open dots and closed dots. what does that symbolize in limits?

Open dots means it doesn't include that point and closed dots mean it does include that point. For example, a open dot at 6 means it can't be 6 but it can be 5.999999 or 6.000001, just not 6. A closed circle for the point 6 would mean it includes 6.

what is the difference between APcalculus AB and APcalculus BC?

More specific course content is given on the College Board website. Essentially, AB is equivalent to Calc 1, while BC is equivalent to Calc 1 and 2. AB covers limits, derivatives, and integrals. BC covers everything that AB does, in addition to derivatives of vector-valued functions, polar functions, parametric functions, planar motion, Euler's Method, improper integrals, integration by parts, arc length, polar areas, the logistic model, and (a whole unit on) series. Hope that I helped.

Main content

Precalculus

Course: Precalculus > Unit 10

Lesson 1: Defining limits and using limit notation

Limits intro

Google Classroom

Limits describe how a function behaves near a point, instead of at that point. This simple yet powerful idea is the basis of all of calculus.

To understand what limits are, let's look at an example. We start with the function

f (x) = x + 2

The limit of

f

x = 3

is the value

f

approaches as we get closer and closer to

x = 3

. Graphically, this is the

y

-value we approach when we look at the graph of

f

and get closer and closer to the point on the graph where

x = 3

For example, if we start at the point

(1, 3)

and move on the graph until we get really close to

x = 3

, then our

y

-value (i.e. the function's value) gets really close to

5

Similarly, if we start at

(5, 7)

and move to the left until we get really close to

x = 3

, the

y

-value again will be really close to

5

For these reasons we say that the limit of $f$ ‍ at $x = 3$ ‍ is $5$ ‍.

You might be asking yourselves what's the difference between the limit of

f

x = 3

and the value of

f

x = 3

, i.e.

f (3)

So yes, the limit of

f (x) = x + 2

x = 3

is equal to

f (3)

, but this isn't always the case. To understand this, let's look at function

g

. This function is the same as

f

in every way except that it's undefined at

x = 3

[What kind of a function is g anyway?]

Just like

f

, the limit of

g

x = 3

5

. That's because we can still get very very close to

x = 3

and the function's values will get very very close to

5

So the limit of

g

x = 3

is equal to

5

, but the value of

g

x = 3

is undefined! They are not the same!

That's the beauty of limits: they don't depend on the actual value of the function at the limit. They describe how the function behaves when it gets close to the limit.

Problem 1

This is the graph of

h

What is a reasonable estimate for the limit of $h$ ‍ at $x = 3$ ‍?

We also have a special notation to talk about limits. This is how we would write the limit of

f

x

approaches

3

\begin{aligned} "The limit of … " & " … the function f … " \\ ↘ & ↙ \\ lim_{x \to 3} & f (x) \\ ↗ \\ " … as x approaches 3 ." \end{aligned}

The symbol

lim

means we're taking a limit of something.

The expression to the right of

lim

is the expression we're taking the limit of. In our case, that's the function

f

The expression

x \to 3

that comes below

lim

means that we take the limit of

f

as values of

x

approach

3

Problem 2

This is the graph of

f

What is a reasonable estimate for $lim_{x \to 6} f (x)$ ‍ ?

Problem 3

Which expression represents the limit of $x^{2}$ ‍ as $x$ ‍ approaches $5$ ‍?

In limits, we want to get infinitely close.

What do we mean when we say "infinitely close"? Let's take a look at the values of

f (x) = x + 2

as the

x

-values get very close to

3

. (Remember: since we're dealing with limits we don't care about

f (3)

itself.)

$x$ ‍	$f (x)$ ‍
$2.9$ ‍	$4.9$ ‍
$2.99$ ‍	$4.99$ ‍
$\underset{close to 3}{\underset{⏟}{2.999}}$ ‍	$\underset{close to 5}{\underset{⏟}{4.999}}$ ‍

We can see how, when the

x

-values are smaller than

3

but become closer and closer to it, the values of

f

become closer and closer to

5

$x$ ‍	$f (x)$ ‍
$3.1$ ‍	$5.1$ ‍
$3.01$ ‍	$5.01$ ‍
$\underset{close to 3}{\underset{⏟}{3.001}}$ ‍	$\underset{close to 5}{\underset{⏟}{5.001}}$ ‍

We can also see how, when the

x

-values are larger than

3

but become closer and closer to it, the values of

f

become closer and closer to

5

Notice that the closest we got to

5

was with

f (2.999) = 4.999

and

f (3.001) = 5.001

, which are

0.001

units away from

5

We can get closer than that if we want. For example, suppose we wanted to be

0.00001

units from

5

, then we would pick

x = 3.00001

and then

f (3.00001) = 5.00001

This is endless. We can always get closer to

5

. But that's exactly what "infinitely close" is all about! Since being "infinitely close" isn't possible in reality, what we mean by

lim_{x \to 3} f (x) = 5

is that no matter how close we want to get to

5

, there's an

x

-value very close to

3

that will get us there.

If you find this hard to grasp, maybe this will help: how do we know there are infinite different integers? It's not like we've counted them all and got to infinity. We know they are infinite because for any integer there's another integer that's even larger than that. There's always another one, and another one.

In limits, we don't want to get infinitely big, but infinitely close. When we say

lim_{x \to 3} f (x) = 5

, we mean we can always get closer and closer to

5

Problem 4

$x$ ‍	$g (x)$ ‍
$- 7.1$ ‍	$6.32$ ‍
$- 7.01$ ‍	$6.1$ ‍
$- 7.001$ ‍	$6.03$ ‍
$- 6.999$ ‍	$6.03$ ‍
$- 6.99$ ‍	$6.1$ ‍
$- 6.9$ ‍	$6.32$ ‍

What is a reasonable estimate for $lim_{x \to - 7} g (x)$ ‍?

Another example: $lim_{x \to 2} x^{2}$ ‍

Let's analyze

lim_{x \to 2} x^{2}

, which is the limit of the expression

x^{2}

when

x

approaches

2

We can see how, when we approach the point where

x = 2

on the graph, the

y

-values are getting closer and closer to

4

We can also look at a table of values:

$x$ ‍	$x^{2}$ ‍
$1.9$ ‍	$3.61$ ‍
$1.99$ ‍	$3.9601$ ‍
$\underset{close to 2}{\underset{⏟}{1.999}}$ ‍	$\underset{close to 4}{\underset{⏟}{3.996001}}$ ‍

$x$ ‍	$x^{2}$ ‍
$2.1$ ‍	$4.41$ ‍
$2.01$ ‍	$4.0401$ ‍
$\underset{close to 2}{\underset{⏟}{2.001}}$ ‍	$\underset{close to 4}{\underset{⏟}{4.004001}}$ ‍

We can also see how we can get as close as we want to

4

. Suppose we want to be less than

0.001

units from

4

. Which

x

-value close to

x = 2

can we choose?

Let's try

x = 2.001

{2.001}^{2} = 4.004001

That's more than

0.001

units away from

4

. Alright, so let's try

x = 2.0001

{2.0001}^{2} = 4.00040001

That's close enough! By trying

x

-values that are closer and closer to

x = 2

, we can get even closer to

4

In conclusion,

lim_{x \to 2} x^{2} = 4

A limit must be the same from both sides.

Coming back to

f (x) = x + 2

and

lim_{x \to 3} f (x)

, we can see how

5

is approached whether the

x

-values increase towards

3

(this is called "approaching from the left") or whether they decrease towards

3

(this is called "approaching from the right").

Now take, for example, function

h

. The

y

-value we approach as the

x

-values approach

x = 3

depends on whether we do this from the left or from the right.

When we approach

x = 3

from the left, the function approaches

4

. When we approach

x = 3

from the right, the function approaches

6

When a limit doesn't approach the same value from both sides, we say that the limit doesn't exist.

Problem 5

This is the graph of function

g

Which of the limits exists?

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boddhulas
Posted 6 years ago. Direct link to boddhulas's post “In the last question, how...”
In the last question, how does
x→7
lim
g(x)
exist? It has two locations right?
Button navigates to signup pageComment on boddhulas's post “In the last question, how...”
(113 votes)
Answer
- Rachel
  Posted 6 years ago. Direct link to Rachel's post “The limit exists because ...”
  The limit exists because the same y-value is approached from both sides. It does not have two locations because the open circle is a just gap in the graph. The closed circle is the actual y-value for when x=7.
  Comment on Rachel's post “The limit exists because ...”
  (236 votes)
luiza
Posted 5 months ago. Direct link to luiza's post “first day learning calcul...”
first day learning calculus :D
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(66 votes)
Answer
- jaydenthegreat1314
  Posted 5 months ago. Direct link to jaydenthegreat1314's post “Same”
  Same
  Comment on jaydenthegreat1314's post “Same”
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K M
Posted 6 years ago. Direct link to K M's post “In problem 5 why can one ...”
In problem 5 why can one of the answers be x→6? but not x→3?
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(20 votes)
Answer
- Dave
  Posted 4 years ago. Direct link to Dave's post “I assume you are talking ...”
  I assume you are talking about the last example. As you approach x=6 from the left you move closer to 3; AND as you approach x=6 from the right, you also moce closer to 3.
  
  As you move closer to x=3 form the left you move closer to 3, BUT when you move closer to x=3 from the right, you move closer to 6.
  
  They must be moving to the same value of y from both sides if you are not going to specify the side in th limit notation.
  Button navigates to signup page
  (25 votes)
Amanda Lu
Posted 6 years ago. Direct link to Amanda Lu's post “For the last question, ho...”
For the last question, how is there a limit for x-->6? There is a point there. I thought there could only be limits if there were open dots.
Button navigates to signup pageButton navigates to signup page
(8 votes)
Answer
- Stephen.Burke
  Posted 6 years ago. Direct link to Stephen.Burke's post “As Sal explained both in ...”
  As Sal explained both in the video Limits intro, and in the text, the beauty of limits, and one property of limits, is that they do not explain the actual point of the graph, but the behavior leading up to that point. Whether there is a point at f(x)= 6 or a hole, that would not change that there still is a limit, unless a jump occurs between the two leading lines.
  Comment on Stephen.Burke's post “As Sal explained both in ...”
  (32 votes)
sam
Posted 5 months ago. Direct link to sam's post “learning this calculus co...”
learning this calculus concept here was actually super fun. although i know this will take many hours, the hours will be exhilarating! i'm going to try and conquer calculus in 2 weeks. (today is june 24th, i have until july 10th)
Button navigates to signup pageComment on sam's post “learning this calculus co...”
(16 votes)
Answer
- Master Mind
  Posted 5 months ago. Direct link to Master Mind's post “How is calc going so far ...”
  How is calc going so far for ya'll
  Comment on Master Mind's post “How is calc going so far ...”
  (3 votes)
mzhang
Posted 6 years ago. Direct link to mzhang's post “the plot has open dots an...”
the plot has open dots and closed dots. what does that symbolize in limits?
Button navigates to signup pageButton navigates to signup page
(5 votes)
Answer
- Allam Amzad
  Posted 6 years ago. Direct link to Allam Amzad's post “Open dots means it doesn'...”
  Open dots means it doesn't include that point and closed dots mean it does include that point. For example, a open dot at 6 means it can't be 6 but it can be 5.999999 or 6.000001, just not 6. A closed circle for the point 6 would mean it includes 6.
  Button navigates to signup page
  (21 votes)
cr7neymar9
Posted 7 months ago. Direct link to cr7neymar9's post “over a year ago is was on...”
over a year ago is was on algebra 1 videos now I'm on calculus finally. lets goo
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(15 votes)
Answer
- mezomaxim516
  Posted 6 months ago. Direct link to mezomaxim516's post “Bro what? Congratulations...”
  Bro what? Congratulations, but be sure you have all the fundamentals down from Algebra 1 and 2 along with Geometry and precal. You don't want to jump ahead into something that you don't understand. Be sure to have strong foundations before continuing.
  Comment on mezomaxim516's post “Bro what? Congratulations...”
  (1 vote)
1249424990bsq
Posted 5 years ago. Direct link to 1249424990bsq's post “what is the difference be...”
what is the difference between APcalculus AB and APcalculus BC?
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(5 votes)
Answer
- Alex
  Posted 5 years ago. Direct link to Alex's post “More specific course cont...”
  More specific course content is given on the College Board website.
  
  Essentially, AB is equivalent to Calc 1, while BC is equivalent to Calc 1 and 2. AB covers limits, derivatives, and integrals. BC covers everything that AB does, in addition to derivatives of vector-valued functions, polar functions, parametric functions, planar motion, Euler's Method, improper integrals, integration by parts, arc length, polar areas, the logistic model, and (a whole unit on) series. Hope that I helped.
  Comment on Alex's post “More specific course cont...”
  (17 votes)
Manaf01
Posted 4 months ago. Direct link to Manaf01's post “first day learning calcul...”
first day learning calculus
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(11 votes)
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curiousfermions
Posted 4 years ago. Direct link to curiousfermions's post “When we take a limit, we ...”
When we take a limit, we approach a specific x value from both sides. But there is an infinite number of numbers between any two numbers. So when we are approaching x, can we ever really get there? And if we can not, then does the calculus give us an answer with really really really really really small flaw that does not create problem in any calculation? Or is it 100% flawless?
Button navigates to signup pageButton navigates to signup page
(6 votes)
Answer
- Moon Bears
  Posted 4 years ago. Direct link to Moon Bears's post “This is the issue that a ...”
  This is the issue that a lot of people had in the development of the calculus. The solution was quite clever, the idea is that you can get "arbitrarily close" that is given any epsilon positive there's a delta such that |x-a| < delta implies |f(x) - f(a)| < epsilon. If this is true FOR EVERY epsilon, then we say the limit exists and the function is continuous. Is it 100% flawless? Not really, but the thing is it's useful. Using these ideas we were able to build up a foundation for calculus which eventually lead to the physics that put man on the moon. The point is it gets results.
  Comment on Moon Bears's post “This is the issue that a ...”
  (6 votes)