This channel deserves way more attention. That was such a long proof yet so easy to understand from start to finish. One of those KZread videos teachers will show students because it's better than their own explanation.

@MasterWuMathematics

8 ай бұрын

Thank you very much for your kind words. Yes, I feel that I deserve more attention, but I need to make more videos like this one to achieve it.

@barendtfernandez7533

4 ай бұрын

As "useful" as Monet's water Lilies!

3 years later and this is still helping people, this video helped me understand this limit after watching multiple videos and not understanding , Wish more growth to your channel. hope my sub will be the start of many more.

Can't see this too many times --- thank you for the nice reminder!

Excellent presentation

Great explanation-instant like and subscribe!

@MasterWuMathematics

8 ай бұрын

Awesome, thank you!

Nice explanation!

Nice proof,,,thanks

I do it with lim[(lnx-lnx0)/x-x0] x->x0. becomes limln(x/x0)/(x-x0) x->x0 Then i pose u=x/x0 and the limits becomes lim lnu/[x0(1-u)] that lim(lnu)/(u-1)=1 using N-L theorem and squezze theorem u->1

Awesome You make it so easy more than my teacher does Tks for this

@MasterWuMathematics

8 ай бұрын

You're welcome 😊

woww so amazing

Can anyone explain why in 5:25 you can take the 1/x out of the limit, saying it's "independent of the letter m" - after all m is defined as a function of x?

This was beautiful!

@MasterWuMathematics

8 ай бұрын

Thank you!

@MochiClips

8 ай бұрын

@@MasterWuMathematics I'd only ever seen the chain rule way of doing it (dy/dx= 1/(dx/dy) ) which always felt like a cheat! Will show my students this too thank you!

@abhirupkundu2778

3 ай бұрын

goofy thing that a math teacher didn't know this...@@MochiClips

@MochiClips

3 ай бұрын

@@abhirupkundu2778 did you know there's more than one way of doing things? 🤔

Muy bien la explicación.

Great video. But where did you get the formula for e from?? 1:06

@adamlea6339

3 ай бұрын

That formula is one of the definitions of e.

At 6:08 , how do we account for m -infinity has the same result as n -> + infinity, but is there a way to prove this?

@murphy2k501

8 ай бұрын

the definition of e provides that we should pick +infinity. If we pick -infinity one, the equation can't be about e. It will be about a number equal to ~ -0.71.

Amazing

thx

Did you need the defn of e?? Your last limit (1+1/n)^n as n approaches infinity is 1, no?

@MasterWuMathematics

7 ай бұрын

Yes… and no that limit does not equal 1, it approaches e. I will demonstrate that in a future video.

beautiful

you can prove this in four steps using implicit differentiation and definition of natural logarithm.

@MasterWuMathematics

6 ай бұрын

What is the proof of the definition of the natural logarithm?

@timm1328

6 ай бұрын

one does not need to prove definitions. one merely needs to state them. the definition is: e^x =y if and only if ln(y) = x

@timm1328

6 ай бұрын

let y=ln(x) then by definition: e^y=x [1]. using implicit differentiation, d(e^y=x)/dx -> By the chain rule: e^y*dy/dx = 1 -> dy/dx = 1/e^y . then by [1] dy/dx = 1/x QED

why did i move 1/x outside the limit of m, when it is actually an expression in terms of m

@MasterWuMathematics

6 ай бұрын

I can be moved outside because it is independent of m.

@BilalAhmed-on4kd

6 ай бұрын

@@MasterWuMathematics but it is an expression in terms of m, so it is DEPENDENT of m

@MasterWuMathematics

6 ай бұрын

But the limit applies to m only, not x.

you didn't explain why we can take the limit inside the square bracket.

@Rabbiddogg-wf1db

Жыл бұрын

It’s not immediately clear but that’s just a limit property

@samarjyoti-ray

Жыл бұрын

​@@Rabbiddogg-wf1db could you please link a source to the limit property?

@Rabbiddogg-wf1db

Жыл бұрын

@@samarjyoti-ray Sure, watch Professor Leonard’s Calc I playlist, specifically Lecture 1.2 Properties of Limits. At the 1:28:00 mark he explains why you can move a limit to the inside of a trig function by composition. Same concept would apply to the natural log function I believe.

@samarjyoti-ray

Жыл бұрын

​@@Rabbiddogg-wf1db ah, I see! if, lim x->a f(g(x)) = f(lim x->a (g(x)), provided lim x->a g(x) exists and is continuous at x=a. thanks mate!

@Rabbiddogg-wf1db

Жыл бұрын

@@samarjyoti-ray Glad to help!

Par définition le logarithme est la primitive de 1/x

@MasterWuMathematics

7 ай бұрын

Yes but how do you get to that definition?

@ammoursidicharef1512

7 ай бұрын

@@MasterWuMathematics le logarithme est la surface de la courbe 1/x entre 1 et x ,puis toutes les propriétés viennent de là et même la réciproque Exp(x) et la valeur approchée de e=2,718....... lorsque la surface est 1 .

@MasterWuMathematics

6 ай бұрын

@@ammoursidicharef1512 I'm not sure I understand. Are you able to write this is in English?

THANKS SO MUCH I SUBSCRIBED PLEASE REPLY SIR!!!

@MasterWuMathematics

8 ай бұрын

Thank you for subscribing. How can I help?

Another approach: Define ln(x) = int(1/t, t=1..x); d(ln(x))/dx=1/x by FTOC. All properties of ln and the exponential function are derived from this starting point.

@MasterWuMathematics

8 ай бұрын

Yes, but you have to know the result in the first place.

This proof is incomplete, since you did not show that the limit as n aproaches infinity of (1+1/n)^n = e but just took that for granted. It is not obvious that this limit converges in the first place, since lim n->infinity [f(n)]^[g(n)], where lim n -> infinity f(n) = 1 and lim n -> infinity g(n) = infinity is an indeterminate form.

@MasterWuMathematics

8 ай бұрын

I'm very sorry but I'll have to show that this limit is equal to e in another video. I'll do that soon. However, you can't apply the limit as you have written here. There is no such limit law that lim n->∞ [f(n)]^[g(n)] = [lim n-> ∞ f(n)] ^ [lim n-> ∞ g(n)]. As you have shown, you get an indeterminate limit, but it's not a valid way to apply a limit.

@weinsterle1999

8 ай бұрын

​@@MasterWuMathematics Until now I have seen two ways of how to prove this limit. The first is using L'Hôpital's rule, which is just circular reasoning; the derivative of the logarithm is 1/x because of the limit and the limit equals e because of the derivative of the logarithm. The second is kind of a weird approach where you start by defining the logarithm as the integral of 1/t from 1 to x. I suppose this works as a proof but it still kind of feels like cheating, since you knew the derivative of the logarithm beforehand, otherwise you couldn't have come up with this definition. I'm interested to see your approach on this.

@orage8802

8 ай бұрын

​@@weinsterle1999but isn't e literally just defined as that limit? Wouldn't trying to prove that limit be akin to trying to prove that pi is the ratio between a circle's circumference to its diameter

@MasterWuMathematics

7 ай бұрын

@orage8802 You're on the right track with that. Euler's number e has a long history and unlike Pi, e has no geometric basis! I'm not aware of any "proof" but this limit was first discovered by Bernoulli when he was studying Compound Interest. I'll do a video on that in the near future.

I know this is a late comment, but a lot of the algebra in this video is unclear, at least to me, it does not seem to be "first principles" as described.

@MasterWuMathematics

8 ай бұрын

How can I help you to better understand? What in particular is unclear?

@ashmain2269

8 ай бұрын

​@@MasterWuMathematicsNear the very end, you say the we should let n = 1/m, where m -> 0, and n -> to infinity if m -> 0, but that in it of itself is a limit, how are we then to plug in that limit into our limit we are trying to resolve? Do we just assume that since we are solving for a limit we can rewrite the limit of m -> 0 as the limit of n -> infinity, since they are both limits?

@MasterWuMathematics

8 ай бұрын

Yes, that is correct, and that is what I’ve tried to demonstrate here. Btw… by “first principles” does not mean I’m taking you back to the beginning. It means using the definition of the derivative in calculus to prove the known result. And the process can be quite challenging. Otherwise we’re just taking it for granted that d/dx lnx = 1/x

Slow as it should be