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**

To find the lt {sin(pi/3+h) - sinpi/3]/h as h--> 0.

We know that Lt {f(x+h)-f(x)}/h and f'(x) are the same.

Therfore we find the value of d/dx (sinx) at x = pi/3.

d/dx(sinx) = cosx.

Therefore {d/dx(sinx) at x= pi/3 } = cos pi/3 = 1/2.

Therefore Lt {sin(pi/3 +h ) - sinpi/3}/h = {d/dx (sinx) at x= pi/3}

Lt h--> 0 {sin(pi/3 +h ) - sinpi/3}/h = cos (pi/3)

Lt h-->0 {sin(pi/3 +h ) - sinpi/3}/h = 1/2.