Calculus

[sweet sphinx]

HOW TO SOLVE THE THIRD DAMN QUESTION

[lean plover]

I'm going to guess this translates to: show that, just because the derivative of a function exists at a point in its domain, that you can't say those limits exist

[sweet sphinx]

they want an example that prove that even if the two limits are equals it doesnt mean that the derivative of a function exisits at a point on its domain

[lean plover]

Oh

[sweet sphinx]

any idea of an example?

[lean plover]

Idk

[plain marsh]

Je me demande si il faudrait pas partir de la définition de la dérivée par la limite avec lim (f(a+dx)-f(a))/dx=f’(a)
Lorsque dx tend vers zéro.

[lean plover]

Maybe like lnx?

Or ln|x|

Idk

[sweet sphinx]

Ofc u would use it

[sweet sphinx]

But they want an example of a function that proves this

[sweet sphinx]

I didnt take this in school yet so idk

[lean plover]

Wait, I think I got it

||Let g(x)=f'(x). What kind of g satisfies this?||

[sweet sphinx]

uh

i dont understand

[sweet sphinx]

yep

But we need an example

[plain marsh]

Exponential function

[sweet sphinx]

Didnt take this yet in school

[lean plover]

Great, any function satisfying f'(x)=x/x

[charred cypress]

maybe overcomplicated but $f(x) = x \sin (\frac{1}{x^2})$ with $f(0) = 0$

[proper cradle]

bloubbloub

[sweet sphinx]

Oh?

Lemme see

[sweet sphinx]

a is 0?

[charred cypress]

yes

[sweet sphinx]

f'(×) exist thi

Tho

Doesnt it?

[charred cypress]

hm actually it doesn't work for the reason you said

[sweet sphinx]

no wait

It does exist

F'(×) exist

and we dont want that

[charred cypress]

it has no limit at zero unfortunately

[dusk furnace]

but what is it that they ask you written in English

[sweet sphinx]

.

[sweet sphinx]

@charred cypress idk if u can understand this but f'(x) exist

[charred cypress]

no because sinx/x goes to 1 when x goes to 0, but 1/x^2 goes to infinity

[sweet sphinx]

Exactly

So f'(×) exists

We dont want it to be derivative on a

[charred cypress]

look at your second line

f'(0) doesn't exist

but f' has no limit at 0 either

[sweet sphinx]

huh

i dont understand

f'(×) goes to infinity

• infinity

whats wrong w my second line

[charred cypress]

you seem to write $\frac{\sin (1/x^2) }{1/x^2} \rightarrow 1$

[proper cradle]

bloubbloub

[sweet sphinx]

No

No

[somber otter]

im sorry but 1/x^2 don't go to 0 asw ? when x -> + infty

[sweet sphinx]

x ---> 0

[somber otter]

oh

mb

[sweet sphinx]

I basically multiplied by 1/x²

[charred cypress]

well that's what I'm trying to tell you: this doesn't work because 1/x^2 goes to infinity

whatever you do, sin(1/x^2) doesn't converge sooo...

[sweet sphinx]

yep yep

[charred cypress]

ok I know

$f(x) = \sqrt{|x|}$

[proper cradle]

bloubbloub

[sacred crater]

What if f(x)= x if x=/0 and 1 if x=0

[sweet sphinx]

the two limits aint equal

[charred cypress]

shit

[sweet sphinx]

1 and -1

[sweet sphinx]

elaborate

But f(×) = x doesnt work 100%

Cuz f'(×) exist

[charred cypress]

well since f is not continuous you could just do that

[sweet sphinx]

The two limits are equal

Wait nvm it doesnt mean that its continuous

[charred cypress]

but for a continuous example I believe $f(x) = x^{1/3}$ also works

[proper cradle]

bloubbloub

[charred cypress]

I should be saying $f(x) = x^{1/3}$ if $x > 0$ and $f(x) = - (-x)^{1/3}$ otherwise

[proper cradle]

bloubbloub

[bright garden]

$f(x) = \begin{cases} x^2 & \quad x<0 \ x^2 + 1 & \quad x \geq 0 \end{cases}$. For this function we have that $f'(x) = 2x$ is defined for every $x\neq 0$, and that $\lim_{x\to 0^{-}} 2x = \lim_{x\to 0^{+}} 2x = 0$. But the function $f$ can not be derivable on $x=0$ because $f$ is not continuous at 0.

[proper cradle]

César

[sweet sphinx]

yes

tysm