Question

without using l'hopitals rule.

limit of h goes to 0 of [ sin 2((3*pi/8)+h)) - sin (3*pi/4)] / h

Answer 1

It looks like a typo to me. If you're trying to find the derivative of sin(x) at x = 3π/4, then you want
[sin(3π/4 + h) - sin(3π/4)]/h
= [sin(3π/4)cos(h) + cos(3π/4)sin(h) - sin(3π/4)]/h
= [(1/√2)cos(h) - (1/√2)sin(h) - 1/√2]/h
= 1/√2 [cosh-sinh-1]/h
= 1/√2 (cos(h)-1)/h - sin(h)/h
as h→0, this becomes just -1/√2 sin(h)/h
I assume you have learned already that the limit of sinx/x = 1, so the final answer is
-1/√2