What is the limit of the derivative of sin x + cos x at x = pi?

[justapurpose]

Homework Statement

Given y = sin x + cos x , evaluate limit h approaches 0 of [(f(pi+h)-f(pi)]/h

Homework Equations

d/dx (sin x) = cos x
d/dx (cos x) = -sin x

The Attempt at a Solution

This was a question on my quiz today. Its answer, which I knew from the glance, had to be the derivative of the equation at x = pi.
Using d/dx rule,
dy/dx = cos x - sin x
at x = pi; cos pi - sin pi = -1 - 0 = -1
I found the answer by using derivative rules but I couldn't do it using the limit concept.

lim h->0 [(sin pi+h)+(cos pi+h)-(sin pi)-(cos pi)]/h
lim h->0 [(sin pi)(cos h)+(cos pi)(sin h)+(cos pi)(cos h)-(sin pi)(sin h)-(sin pi)-(cos pi)]/h
lim h->0 [0-(sin h)-(cos h)-0-0+1]/h
lim h->0 [1-(sin h)-(cos h)]/h

That's as far as I've got. I couldn't cancel the h in the denominator. Please help, thank you.

 

[lurflurf]

[1-(sin h)-(cos h)]/h=[1-cos h]/h-sin h/h
so consider
lim_h-> [1-cos h]/h=cos'(0)=0
lim_h->0 sin h/h=sin'(0)=1
So to avoid using derivatives you will need to find those limits another way in fact trigonometric identities can be used to avoid lim_h-> [1-cos h]/h which leaves
lim_h->0 sin h/h=sin'(0)=1