Question

Using the fact that lim sin(h)/h =1 h->0
and lim cos(h)-1/h =0 h->0
complete the following limits
a. lim sin(x+h)-sin(x) all divided by h h->0

b lim cos(x+h) - cos(x) all divided by h h->0

Answer 1

I sincerely how you mean [cos(h) - 1] / h.

you will have a very hard time demonstrating that cos(h) - 1/h approaches 0.

you must remember your Order of Operations.

Try multiplying numerator and denominator by [cos(h) + 1]. See if something magic pops out.

Answer 2

Identities:
\[\sin(x+h)=\sin x\cos h+\cos x\sin h\\
\cos(x+h)=\cos x\cos h-\sin x\sin h\]

Answer 3

Ok I can rewrite it better:
Using the fact that lim h->0 sin(h)/h =1 and lim h->0 cos(h)-1/h =0

compute the limit for each
a. lim h->0 sin(x+h)-sin(x)/h

b lim h->0 cos(x+h)-cos(x)/h

Answer 4

You did not write it better. You still have not remembered your Order of Operations.

Try this:

Using the fact that lim h->0 sin(h)/h =1 and lim h->0 [cos(h)-1]/h =0

compute the limit for each
a. lim h->0 [sin(x+h)-sin(x)]/h

b lim h->0 [cos(x+h)-cos(x)]/h

The brackets are NOT optional.