Evaluate lim [sin(pi/3+ h) - sin(pi/3)] / h as h---> 0

lim [sin(pi/3 + h) - sin(i/3) / h  as h-->
0

It is obvious frm the definition of the derivtive. we
know that:

f'(x)= lim (f(x+h) - f(x)]/h   as h -->
0

Thenwe will assume
that:

f(x) =
sinx   

==> f'(pi/3) = lim [sin(pi/3 + h) - sinpi/3]
/h  as h-->0

Now we will differetiate
f(x)

f'(x) = cosx

==>
f'(pi/3)= cos(pi/3)

                   
=cos60

                     =
1/2

==>lim [sin(pi/3 + h) -
sin(pi/3)]/ h   as -->  is 1/2