Question

lim [cos(pi/2 +h) - cos(pi/2)]/h
h-->0

Answer 1

use the addition angle formula for cosine

Answer 2

you can use the identity :
cos(pi/2 +h) = -sin(h)

and dont worry, that cos(pi/2) = 0

Answer 3

so, the limit becomes
lim h->0 -sin(h)/h = ... ?

Answer 4

huh?

Answer 5

how can you get there using a theorm like l'hopitals or trapezoid or soemmthign along those lines?

Answer 6

this question just basic, we neednt l'hopital rule..
just defenition that :
lim h->0 sin(x)/x = 1 and
lim h->0 tan(x)/x = 1

Answer 7

so how do we solve it?

Answer 8

lim h->0 -sin(h)/h
= - lim h->0 sin(h)/h
= - (1)
= -1

Answer 9

where did the sin(h) come from?

Answer 10

Use the definition of the derivative to get
\[
\lim_{h->0} \frac {f(a+h)-f(a)}h= f'(a)\\
\text { with }\\
a=\frac \pi 2\\
f(x)=\cos(x)\\
f'(x)=-\sin(x)\\
f'(a)=-\sin(\pi/2)=-1
\]

Answer 11

The limit is -1

Answer 12

I apologize but I am still confused. Can you help me plug it in because I don't understand what you are doing?

Answer 13

lim [cos(pi/2 +h) - cos(pi/2)]/h
h-->0