banned-users

Finally, when I think I am at the peak, I start to fall

just to lip up and stop forever

Pikachu really is an edgelord

what a tragic story Kappa

sin(x^2)tan(|x|)cos(√x)+asin(√(|x|)-9) is my waifu

she has all the best curves

Back in my day, we didn't stick an audio wave on a graph and call it a function

Did you call it a waifu?

Back in your day, f(x) was considered conventionally sexy

times change

and yes

s(x^2)t(|x|)c(√x)+as(√(|x|)-9)-sama is my waifu

Kappa

Well

that is enough shitposting

I need sleep

tangent is the ratio between the sin and cosine of an angle

well I mean I learned that

are you still dieing to homework @legorhin

not really

my group project that i pretty much soloed was done today

you cant have a limit that is an imaginary number right?

rip leg

why not?

because I dont know

I dont know how to calculus

you can

rip

I have no idea how to answer this then

but your math classes will pretend you can't unless you go into mathematics or electrical/computer engineering

"explain why you can not use substitution to find the limit" lim of rad(x-2) as x approaches -2

the implicit scope of your class is to deal with only real numbers

"forget" about i until later

but

we learned that in algebra

I dont afraid of no is

you are nowhere near the level of math that will actually be finding limits in the complex domain

oh

or doing complex calculus

so we dont give no shits about no is

checked the answer key

what is rad(x)?

(x)^(1/2)

sqrt...

oh

yeah

that isnt really descriptive

rad could be a couple things

idk where I thought of that

yeah it really could be

what is it called when you get 0/0 through substitution?

L'Hospital's rule?

ok google L'Hospital's rule

http://tutorial.math.lamar.edu/Classes/CalcI/LHospitalsRule.aspx

indeterminate form

so (sin(x))/(-2(3.14...)</=x<0) does not work in my calculator

rip

how would I go about doing this @legorhin

what?

essentially what I am trying to do is limit the function to (-2(3.14...)</=x<0)

what is </=

(sin(x))/(x<0) works

that is less than or equal to

<= is the programmers way of showing that

ok

is it >= or =>

out of curiosity

>=

double character operators with = in them almost always end with it

anyway

how would I do that

sin(x)/2π <= x < 0

sin(x) <= 2πx < 0

well

the first one does not show up at all

rip

sin(x)/x <= 2π < 0

disproven

sinc(x) <= 2π < 0