How the fuck do you solve this Sup Forums?

How the fuck do you solve this Sup Forums?

I know:

(1-cosx)/x = 0

but how do I get there???

its insoluable

insoluble*

Also book sais its is solvable.

idiots
the answer is 3^2²(sin*x²)

...

prove me right faggot

its easy af: sin(x)/x goes to 1 and you said you know (1-cosx) goes to 0 for x to 0. So 1 *0 =0

...

well, just saying it can never actually equal zero. My guess would be DNE because as you approach from the left it is positive infinite and from the right it is negative infinite. I think.

Use Lopital's rule/theorem whatever.
Do the derivative of sinx(1-cosx) and the derivative of x^2. but separatly.

Then make a new fraction with the results.

winrar

l'hopital, idiot.

That's the algebraic definition of pie. As in give me ur fuckin' pie nigger before I go full harambe on your bitch ass tear it open like prison rape and then get shot by police. fgt.

Also none of these answers are correct. As x goes to 0 the denominator goes to 0, which ordinarily could imply an infinite but sin and cos are oscillating between -1 and 1, meaning there's either some algebraic answer I can't remember how to find or it's undefined.

the answer is zero
the function will never equal zero but it always approaches zero while it never equals zero.

Yeah, L'hopital's rule. Derivatives of num and den then ratio. Derp.

I'm still bout to rape dat ass nigga tho

What's up with that bigass number 2. It's larger than the x lmao

Just take a ton of Adderal. You'll either get a 4.00 or die. Win either way tbh,

underage b&

Use calc 1 knowledge ... l'hopital's rule

Use Taylor expensions. 1-cos x = x^2/2+O(x^3)
The limit you want is 1/2 sincs sin(y)/y has limit 1 when y->0

What the fuck are you talking about? sinx/x is 0/0 indeterminate, not 1


Use Lopital's wrong you fucking dumbass

I just started cal 1 4 days ago. The book sais the answer is 0.

yeah needs an accurate proof. You can show it on the unit circle. just google it. lim x to 0 sinx/x is actually very common in calc 1 kek. google it

Trust me, i'm going to be an engineer:
lim sinx/x x->0 = 1
so its easy, (sinx*(1-cosx))/x^2= sinx/x * 1-cosx/x
The product of limits its the limit of the product, then:
1 * 0 = 0

senx/x = 1

(sen(1-cosx))/x^2 = (senx/x)((1-cosx)/x)

Exactly, and the end result is 1x1 so 1

L'Hopitals dumbass

It's sinx/x * 1-cosx/x. So that's 1 * 0 = 0

limit -->0 sinx -sinxcosx / x^2

lhopital it
dsinx = cosx dcosx = -sinx
d(sinxcosx) = sinx dcosx + dsinx cosx
= sin^2 x - (-cosx * cosx)
= sin^2x + cos^2x = 1

breaks down to limit -->0 ( 1/2x ) = 0
ZERO


dx^2 = 2x

so limit ->0 of cosx -

the answer is fucking 0

go to symbolab dot com and type it in

Sorry, 1x0 so 0

Use your limit laws faggot

that has to have an error in it, my bad. redoing

Hopital with periodic functions is useless

>Put in graphing calculator
>0.5
Easy

so... sinx - sinx cosx / x^2
per previous derivative calculations....
. l'Hopital's rule

(cosx - 1 ) / 2x

1/2x -> 0

so only cosx/2x left

0.5 x

if I"m wrong, so I'm a faggit....whatever

>Use l'hopital
Ok
lim (x->0) sin(x)/x
= lim (x->0) cos(x)/1
= cos(0)
= 1

0/0=1

ugh.... .5 / x or .5x^-1


geez, I don't miss this shit at all

0/0 is undefined

for what do you want to know that hehehe cheers

It warms my globe to declare that the limit is nonzero, a winner, making limits great again !

retard