Question

What is the lim h->0 cos(3pi/2+h)-cos(3pi/2)/h

Answer 1

I assume you have to solve this using nothing but the limits?

Answer 2

oh i hope not!

Answer 3

you have been asking calculus questions, implicit diff and so on much harder than this

Answer 4

if you have to solve this with nothing but limits, use\[\cos(a+b)=\cos a\cos b-\sin a\sin b\]first

Answer 5

you cannot just plug into the difference quotient (which is what this is called) ever

Answer 6

if you can use all your calculus, recognize this as the derivative of cosine, evaluated at \(\frac{3\pi}{2}\) and you do it in your head

Answer 7

what is the derivative of cosine?

Answer 8

@Fanduekisses did i lose you there? i hope not

Answer 9

ok -sine(3pi/2)

Answer 10

yup

Answer 11

don't forget
\[f'(a)=\lim_{h\to 0}\frac{f(a+h)-f(a)}{h}\] which is exactly what you have, with
\[a=\frac{\pi}{2}\] and sine

Answer 12

oops i mean
\[a=\frac{3\pi}{2}\] but you knew that

Answer 13

i guess it is not quite clear
you have given to you the definition of the derivative at \(\frac{3\pi}{2}\)

Answer 14

you could compute that limit by hand if you like, but you don't have to since you already know the derivative of cosine is - sine

Answer 15

making your answer
\[-\sin(\frac{3\pi}{2})=1\]

Answer 16

thanks, I knew how to solve this at the beginning of the school year, idk what is going on in my head right now :(

Answer 17

i bet this is one of those reviews right?

Answer 18

yes

Answer 19

just to see if you remember that stuff

Answer 20

OOOOOOHHHHHHHHH I just remembered everything lol, or at least comprehend it =D

Answer 21

whew!!!!