It is not unusual in calculus

textbooks to see something– to see problems along the lines

of, prove that the limit as x approaches– I don’t know,

say, as x approaches 1. Let me think of an interesting

function when x approaches 1. Let’s see. 3x times x minus 1

over x minus 1. And you might say, hey Sal, why

did you do this x minus 1 in the numerator and denominator? Well, I did that because

obviously you can’t have a 0 down here. So this function is actually

not defined as x is equal to 1, right? If I had just put a 3x,

this cancels out for all the other values. So this is essentially the

line 3x, except it has a hole at x is equal to 1. And that’s why I did it. Because I want to take the

limit, and it’s interesting to take a limit a to a point where

the function doesn’t exist. But the books will say, hey,

prove that this limit is equal to– well, it would actually

be equal to 3, right? Prove that. And you essentially have

to use the epsilon delta definition of a limit. So to prove this, you’re

essentially proving what this means. And this means– proving

this is the same thing as proving– so you could

just prove that given–. So if you give me some epsilon

greater than 0– and remember, that’s how close you want f

of x to get to the limit. Given some epsilon greater than

0, in order for this to be true, I need to prove that I

can give you– that there exists– that, you know, I can

give you a delta greater than 0. Where as long as the distance

between x and our limit point, x and 1, is less than delta–

remember, we don’t want x to ever get right on top

of the limit point. Because a function might

not be defined there. So the distance between

x and 1 has to be greater than 0, right? That’s what this tells us. So I have to prove, you

give me an epsilon, that I can give you a delta. And as long as x is within

delta of 1– and this is all this is saying– as long as

the distance between x and 1, right? That’s when you subtract

2 things and take the absolute value. That’s just distance. So as long as the distance

between x and 1 is less than delta then the distance between

3x– let me write that up there– the distance between 3x

times x minus 1 over x minus 1, that’s the function, right? The distance between that and

the limit point– and 3– is going to be less

than the epsilon. A lot of calculus text

books, you know, they’ll write it generally. They’re like, the limit of f of

x as x approaches a of equals l, and so this would be a. This would be your f of

x, and this would be your l, right there. But anyway, when someone says

they want you to prove this, they essentially want

you to prove this. I’ll give you an epsilon. You have to prove that any

epsilon that you’re given, or that you give me, that I–

let’s say I’m the one who has to prove it– any epsilon you

give me, I have to be able to give you a delta where, as long

as x is within delta of 1, then the function is within

epsilon of the limit point. And how do we do that? And I’ll just say, you know, a

lot of proofs, it’s just kind of– there’s not a– well,

there’s not a systematic way to do proofs. And that’s kind of what

makes them so interesting on a lot of levels. But these, there kind of tends

to be a systematic way. And this problem I’ve given

you, to some degree, is kind of the easy one that they’ll give

when they ask you to prove these. But they usually involve

starting from this point right here and algebraically

manipulating it until you get– until this expression

right here looks something like this expression. And then you’ll have

this expression is less than, you know, epsilon

divided by something. Or, you know, some

function of epsilon. And it’s like, oh, well as long

as delta is that, you give me an epsilon, I can just apply

that, you know, take epsilon and divide it by

4, or whatever. And use that as my delta. Then I’ve proved the theorem. And so we’ll do that here. So let’s start with this. Let’s start with kind of

where we want to get to. And then figure out a delta

that’s essentially a function of epsilon. So that whenever I’m given an

epsilon, I can say, hey, whatever number you give me,

I’m going to give you back a number that’s, you know,

that divided by whatever. And then this will all work. So let’s do that. So immediately we

can simplify this. We can cross these out. And when you cross that out,

you still have to say, well, we can’t do that as long as x–

you can only do this as long as you assume that x

does not equal 1. Because if x equals 1, this

function is undefined, right? And that’s OK because we only

care about the intervals as x approaches 1. We only care about the

intervals as x approaches 1. We don’t care exactly

when x is equal to 1. So that’s fine. So then this simplifies to–

this right there simplifies to– the absolute value of 3x

minus 3 is less than epsilon. And then we could factor out a

3, so that’s, you know, we could say the absolute value of

3 times x minus 1 is less than epsilon. And you might already see, we

already have an x minus 1 there and an x minus 1 there. If we can get this 3 over

on this side, then we’ll be done, essentially. So the absolute value of 3

times some other number, it’s going to be the same thing as

the absolute value of 3 times the absolute value

of x minus 1. And if you don’t believe me,

I mean, try it out with a bunch of numbers. This is going to be

positive no matter what. This is going to be

positive no matter what. It’s going to have the

same magnitude, right? And of course, that is

less than epsilon. And this is what we say– you

know, a lot of times a calculus teacher will say, if this is

true, this is true, if and only if this is true. And that’s just a

fancy way of saying. So sometimes they’ll

do a two way arrow. Sometimes they’ll write IF

and only IF, kind of an IFF. An IF with 2 f’s. But that just means if

that’s true, that’s true. And if this is true,

then that’s true. And that’s generally the case,

whenever you just do some algebraic manipulation. You can manipulate this back to

get that, or that to get that. And then they’ll say, well, if

this is true, this is true if and only if this is true. But this is just

algebraic manipulation. Right? Because you can go back and

forth between these steps. And that’s true, if and

only if that is true. And then the absolute value

of 3 is just 3, right? We know that. Let’s divide both sides by 3

and you get– well, I can just get rid of the absolute

value signs right there. Because we know that’s 3. If you divide both sides of the

equation by 3, you get the absolute value of x minus 1

is less than epsilon over 3. So what have we

just done so far? We’ve just proved that the

absolute value– I need to write this in another color. We’ve just proved that the

absolute value of essentially 3x times x minus 1 over x minus

1 minus 3 is less than epsilon, if and only if x minus 1 is

less than epsilon over 3. Right? So it’s like, this is true, as

long as the distance between x and 1 is less than

epsilon over 3. So we can just use epsilon

over 3 as our delta. Remember the whole point of

this was like, you give me an epsilon, you give me a

distance, or you say, I want to be within this distance

of my limit point. I want the function to be that

close to the limit point. And this is just the distance

between the function and the limit point, right? This is f of x. And this is the limit

that it approaches. So you want to be that close. And I say, well, you’re going

to be that close if and only if x– the distance between x and

1, or the distance between x and the value it’s

approaching– is less than this. So we just prove

this algebraically. So whatever number you give

me– if you give me a– if you say, Sal, I want to be

within 1 of the limit point. I want f of x to be no further

than 1 from the limit point, then I’ll say, OK, and

I’ll give you 1/3, right? Because this is true. Let me write that out. So let’s say you pick

1 as an epsilon. So you say, hey, you need to

prove to me that there’s some x value where, as long as the x

value is no more– there’s some value of delta. So as long as we’re no more

than delta away from 1, then the function itself will

be no more than 1 away from the limit value. Right? And then I say, well,

you, gave me an epsilon that’s equal to 1. So I could just make– I can

just say this is, you know, we just proved, if and only if

absolute value of x minus 1 is less than whatever number

you gave me divided by 3. And that actually makes

a lot of sense if you graph this equation. Let me see. Let me draw the x-axis, y-axis. We already established. This thing looks just like the

equation of 3x except it has a hole at x is equal to 1. Let me draw the graph. It’ll have slope. It’ll have the slope of 3. So it’ll be pretty steep. It’ll look something like that. And when we’re at 1–

this point right here– there’s a hole. Right? This is 3. So all we’re saying, we just

proved that– so if you said that you wanted to be within 1

of our limit point, that’s like you giving me an epsilon of 1. So this distance

right here is 1. Which would be a

pretty big number. So you want the function

to be within 1 of our limit point, right? Because this is

the limit point. Then all we’re saying is, you

just have to take 1/3 of that. So we just have to be within

1/3– so this distance right here is 1/3 away from 1. Or we could have said it more

generally, this is epsilon. This is epsilon. This would be epsilon over 3. This is epsilon over 3. And the reason why this is a

proof, is because it shows no matter what epsilon you give

me, I can always find a delta. Because I can just take

whatever number you gave me, and divide it by 3. And I can divide any real

number greater than 0. I can divide any

real number by 3. So I can always– no matter

what you give me– I can always find you something else. And therefore, I have proven by

the epsilon delta definition of limits that the limit

as x approaches 1, of this, is equal to 3.

So given an Epsilon, I can give you a range of Delta. But how does that prove the limit actually exist and is true

you rock!!

Man I am glad I looked this up in my 3rd week of freshman year

same dude, FINALLY!!!!

Praise our lord and saviour Khan!

1/3 of E(epsilon)

One third of the Epsilon. To get the delta.

and now I understand 😀

Glad to hear I'm not the only one. I'm a joint honours physics and math and my physics prof used this limit definition in the definition of phase spaces, so I've decided I'd finally go back and make sure I understand this well.

I never understand anything until Khan academy explains it to me.

But in the limit definition it says I FOR ALL X, is it okay to just ignore the fact that x could in fact be 1?

So frustrated. These videos teach you the easiest most basic forms of topics that you can find in any textbook, but never anything beyond that.

OMG exam tomorrow and you saved my life! thank you 🙂

I wish I would have known this before I took the calc test that I failed. Can I just not go to classes and learn from you?

nice and understandable Xplanations… thanx alot

Thank you! This helped so much!

Oh, God! Finally I understood the definition of limits! Thank you Khan 🙂

Thank you!! My hero!

Thank you so much, youre the reason i'll pass calc

Good, but 240p is tough to watch. Please upload your videos to at least 480p.

fucking sal is the man yo!

Why couldn't you be my teacher!

wonderful work. you manage to be precise without being overwhelming and the moderated length (measured in t dimension) of the videos is highly pedagogical.

Thank You ,sir.

I really appreciate you. 😀

You are the "Lionel Messi" of Education

Thank you a lot! That helped me in a way you can never imagine how!

Awesome. Love the notation. Thanks Sal!

YOU ARE A LIFE SAVER!! THANKSSSSS

MILLE GRAZIE

How do you solve for this!?

Lim (x^2+3)=4

X->1

Looks like the ratio is always the slope at the value. Maybe that is only true cuz this is a simple graph

The cleanest explanation I have ever seen.

I'm in the best math university in my country and nobody can explain it as good as you. Thanks!

You are a math hero. Thanks for the help!

does he go into multi-variable calc later on?

Beautiful explaination ! ….Thank you .

Thanks for the video =) it would be really awesome if you uploaded more videos of epsilon delta proofs of limits of more complicated functions (of one and two variables). I'm sure I'm not the only one who would appreciate that very much!

guys, how does he know that the limit is 3?

is this usually in calculus I ?( calculus one )?

THANK YOU.

but doesn't plugging in 1/3 into x give 1, which isn't within 1 of 3 or 3 (L).

I wish we had teachers like you at my college. I never understood this concept and i always ran away from it. But today i totally understood everything just because of your explanation. Thanks a lot for making such a cool site and that too for free! Students like us will always appreciate you for making our studies so much more fun with your cool videos and site. 🙂

The "only if" bit in this talk is irrelevant, really. The definition of limit does not require that. An unfortunate lapse which further obscures that matter of limits which students have problems assimilating anyway. Otherwise, a good video, especially in terms of the instructor's manner of speaking — lively, articulate, accessible. Thank you, colleague.

I am a project manager and work at a company.

I spend my free time with Khan videos when I need relaxation, I sometimes watch before getting on a plane, sometimes at my breakfast table, sometimes prior to sleep.

These videos help me remember all those calculus equations and theorems from my university life. It's like seeing old photos with my friends. I prefer Salman Khan over funny cat videos 🙂

sal i have a question:

At 3:30 why haven't you stated that [3x(x-1)/x-1] -3 also NEEDS to be bigger than zero because if it does equal to zero that means that in the given function at 3 on the y-axis there is a point that exists. HOWEVER

at 3 on the y-axis there shouldn't be any point in existence because if x cannot be equal to 1 y cannot be equal to 3, unless you've defined that there is a point on the y-axis at 3 but not when x=1 (which you haven't done)

Thank you for the awesome video, Sal. Never forget how many students you are helping!

helping people in chile, thanks

Love from Malaysia.

one question. 9:33 how do we come up with the equation of that line with slope of 3?

For any x less than delta and greater than zero, shouldn't the difference between f(x) and L not only be less than epsilon, but also greater than zero? I.e shouldn't it be 0‹|f(x)-L|‹ε

You made this much easier to understand than my college professor.. great job!

Really good explanation…

I thank u for such video u uploaded.

At the beginning of the video, you proved that lim x->1 3x(x-1)/(x-1) =3. Now my question is that if

lim x->a f(x)g(x)/g(x) = f(a) where g(a)=0

if i can give a dollar every time khan academy saved my not focusing in class because my lecturer weak ass …u would make a fortune

if you see that things cancel out, i.e right at the start, can you just cancel them out, insert your a, get your y and be done? that is assuming the question doesn't want proof

haha I love it!

"IFFFFFFFFFF…"

Poor quality

So more simple and correct me if I'm wrong but you take the F of the F(x) function divide epsilon with and then you have Delta.

Thanx so much Sal that was soo helpful 👏😃

I don't know what my education would've been like without Kahn.

Currently taking calc analysis as a com sci major, and this video has proven INVALUABLE, watching it in 2017 8 years after it was made and it's literally doubled my understanding of limits

Good job! Thanks for explaining.

Omg video soo gooood!

Thank You!

You know something is tough even if a great teacher teaches and you still don't understand

lim x → π/2 sin ( x )=1. epsilon delta ?

Using epsilon delta definition of a limit prove that

Lim 1/X = 1/5

X tends to 5

Had been trying to understand this for a week and I was stressing out!! And boom it just clicked! I’m so happy, thanks!

Can't believe this is in Calc 1 for me but that's okay. I feel like I can learn everything from you!

This video is pissing me off royally. I have watched the segment at 2:40+ at least six times and he goes so fast glossing over what is the freaking a?

Going to find a clearer video.

I understand that we start with an epsilon…but why do we start with an epsilon? That never quite made sense to me…

Thanks sir.

True MVP

Thanku very much sir….it clears my concept upto 90 percent..

I liked this explanation

This is is better than the updated video you guys have

Thank you so much for all of your videos. I read and re-read my textbook and reviewed lectures for HOURS on end and still couldn't make heads or tails of the course material. Then I watched a few ten-minute clips from your channel and completely understood what I was doing and how to complete my assignments. Thanks to your videos, I just got 100% on my first quiz.

By the way, I haven't taken a math course in 16 years, and was just thrust into Calculus. Another testament to how good these videos are!

I find it really sad that Salman could explain this to me in 10 minutes and 55 seconds but my professor lost our whole class during an hour and fifteen minute lecture.

Let's just hope my brain doesn't break during the test…

This is not notation friendly at all… my professor would have been very triggered have he seen this hahaha

Is delta=€/3 in this case.

And how can we evaluate the limit of constant and identity functions in terms of epsilon-delta

K now prove a limit doesnt exist

perfect

That moment when it FINALLY made sense. So good. Ty

Everything was so easy before this Epsilon-delta thing

You make math beautiful, sir

Thanks a lot

i love you

This was so helpful

I don't comment on videos like these but I have to, you took a topic I've been struggling with for days and made me understand it so well. I love you homie.

AMEN

Finally this actually makes sense

The best University of the world.

I will donate you when I will earn money.

But you need to wait for at least 5 years

感谢🙏

Thank you Mr. Khan for teaching the entire concept of limits

When the limit aproach to a

Is delta always > |x-a| ?

The video is almost 11 years and still very helpful. That's the beauty of math, the rules never change.

Thank you dude, you have saved me