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

hala718 | High School Teacher | (Level 1) Educator Emeritus

Posted on

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

neela | High School Teacher | (Level 3) Valedictorian

Posted on

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.