Why

[opal adder]

Hiii

Here’s a thing I only accepted but never tried to know why

[gritty bronze]

hmm.. that is what happens with the elementary ones

[opal adder]

I guess you could use:

lim (f(x+h)-f(x))/h when h is close to 0 to prove it. (Using definition of derivative) or you could use the logarithm

[opal adder]

You mean elementary functions?

[gritty bronze]

yes

[wraith meadow]

e^x(h-1)/h

[opal adder]

Yes almost!
It is equal to e^x((e^h)-1)/h

What happens if h is very close to 0

[wraith meadow]

e^x(1-1)/h

[opal adder]

Yes exactly

e^x(0/h)

0/h when h is very close to 0?

Well 0/h is equal to 1

so e^x(1)=e^x

You proved why (e^x)’=e^x

When you want to prove the derivative of a function. Always use that:

[wraith meadow]

I don't believe h -> 0, but not 0

[opal adder]

h is so close to 0 you can consider it is equal to 0

Like really really close

It’s not EXACTLY 0 but very very close

Like 0.00000000000000000000000000000000000000000000000000….1

Do you understand?

[wraith meadow]

No

[opal adder]

It is not exactly 0

It’s not equal to 0 yeah

I meant it’s very close to 0

[wraith meadow]

Counter example lim(x->0) sin x / x = 0/x = 0 but = 1

[opal adder]

Well yeah for sin it is equal to 1

Do you know why?

series expansion

But for exponential,

(e^h-1)/h=1

[wraith meadow]

cos x/1

[wraith meadow]

Why

I don't believe

[opal adder]

This is called the series expansion

This is a way to approximate a function with a polynomial when this function is close to 0

[wraith meadow]

))))))))))))) The Maclaurin series requires a derivative of the function

[opal adder]

Well true. Taylor series is related to derivatives

This is the general formula

When you want to calculate limit for example when you have something like 0/0 and you don’t know the answer

You use the Taylor series approximation

a is h

[wraith meadow]

It remains to prove lim(h->0) e^h-1/h

[opal adder]

Yes and you have to use the series expansion formulas to prove that

Using the series expansion formula for ln(1+x)

[wraith meadow]

]t = e^h -1, h = ln(t+1), t -> 0 lim(t->0) t/ln(t+1)

[opal adder]

Yea exactly

That’s what I was writing

[wraith meadow]

)

[opal adder]

:)))

You are good at this !

[wraith meadow]

And what's next ?

[opal adder]

I’m back

Now you have lim t/ln(t+1) when t->0 right?

You can rewrite it:

lim 1/(1/t)ln(t+1)) when t->0 do you agree?

I just put the t in the denominator

By multiply by 1/t each side

Do you understand?

[wraith meadow]

Yes

[opal adder]

Alright and you know that (1/t)(ln(1+t))=ln((1+t)^1/t))

Do you agree?

Because ln(a^n)=nln(a)

$lim \frac{1}{ln((1+t)^{1/t})};
$ $t->0$

Wait it’s not writing the right way

[analog mortar]

AstroGuy

[opal adder]

Here

Finally!

[wraith meadow]

How did you even come up with that?

[opal adder]

Okay let me show you.

[wraith meadow]

t->0 ln(t+1)/t = 1 => t/ln(t+1) = 1

[opal adder]

$lim \frac{t}{ln(t+1)};$

$lim \frac{1}{(1/t)ln(t+1)};$

$lim \frac{1}{ln((t+1)^{1/t})}$

[analog mortar]

AstroGuy

[opal adder]

Do you understand the steps?

We’re at the end

Of the proof

Almost

[wraith meadow]

1/ln e

[opal adder]

Yes

[wraith meadow]

How did you even come up with that?

[opal adder]

Because lim (1+t)^1/t=e when t->0

That’s… another proof…

So lim ln((1+t)^1/t))=ln(e)=1 when t->0

Basically you need to prove that

….

It’s harder than I thought

[wraith meadow]

e def= lim(n->inf) (1+1/n)^n

]t = 1/n, t->0 lim(t->0) (1+t)^1/t

[opal adder]

Oops wrong one

[wraith meadow]

QED ?

[opal adder]

https://youtu.be/gciJWEVlAuo?si=jgpQK0z7GPxCmZiC

That limit Im talking about

The upper one

Sorry for confusing you

Basically your limit end up with (e^x)’=lim e^x•(1/ln((1+t)^1/t)) when t->0

But you have to prove that lim (1+t)^1/t=e when t->0

Once you prove that, you win

[wraith meadow]

? t->0+ this

[wraith meadow]

n -> +inf, ]t = 1/n => t ->0+

[wraith meadow]

?

[opal adder]

Remember you had:
(e^x)’= lim e^x(e^h-1)/h when h->0 right?

[wraith meadow]

yes

[opal adder]

Then you stated e^h-1=t

So h=ln(t+1)

(e^x)’=lim e^x(t/ln(t+1)) when t->0

[wraith meadow]

?

[opal adder]

I just replaced (e^h-1) by t and h by ln(1+t) in the limit

[wraith meadow]

next

[opal adder]

Oh okay and then i multiplied by 1/t above and under the fraction

It had given:
(e^x)’=lim e^x(1/(1/t)ln(1+t)) when t->0

And with log rule: (1/t)ln(1+t)=ln((1+t)^1/t))

So it becomes:

(e^x)’=lim e^x(1/ln((1+t)^1/t)) when t->0

And by definition lim (1+t)^1/t=e when t->0

So you get (e^x)’=e^x(1/ln(e))

Hence (e^x)’=e^x

[wraith meadow]

yes

[opal adder]

You just have to prove that

To prove (e^x)’=e^x, you have to prove lim (1+t)^1/t=e when t->0

[wraith meadow]

t->0+

[opal adder]

Yeah 0+

[wraith meadow]

.

[opal adder]

Positive zero

[wraith meadow]

e def = lim(n->inf) (1+1/n)^n

[opal adder]

Oh yes you’re right!

[wraith meadow]

n->+inf

[opal adder]

You used infinity limit to prove the limit when t->0

[wraith meadow]

]t = 1/n => t->0+

lim (t->0+) (1+t)^1/t

= e

[opal adder]

Yes

Wow now I understand why they used the infinity limit

You’re clever

[wraith meadow]

Thank you

[opal adder]

Did you understand how to prove (e^x)’=e^x?

It all starts with this formula

[wraith meadow]

yes

[opal adder]

Alright!

Can I add you?

[wraith meadow]

yes

[opal adder]

Thanks!

I sent you the friend request

[wraith meadow]

t->0+

[wraith meadow]

t->0

lim (t->0-) (1+t)^1/t, ]m = -1/t, t = -1/m, m->+inf, lim (m->+inf) (1-1/m)^-m = lim 1/((m->+inf) (1+(-1)/m)^m) = 1/lim (m->+inf) (1+1/m)^-1 = 1/e^-1 = e