lim (sin(x+h)-sin(x))/h
h→0
since sin(x+h) = sin(x) cos(h) + cos(x) sin(h)
then we have
lim (sin(x) cos(h) + cos(x) sin(h) - sin(x))/h
h→0
grouping 1st and 3rd term, we have
lim ((cos(h) - 1) sin(x) + cos(x) sin(h))/h
h→0
Since h not equal 0, a limit exist for the above function.
When h→0 cos(h)=1 and sin(h)=h, the above function becomes
lim ((1-1)sin(x) + cos(x)(h))/h
h→0
lim (0 + cos(x)h)/h
h→0
lim cos(x)h/h
h→0
cos(x)
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equation used in this exercise:
sin(a+b) = sin(a)cos(b) + cos(a)sin(b)
and
lim cos(x) = 1 ; lim sin(x) = x
x→0 x→0
applying equations and recognizing pattern of problem is key to solving this kind of math questions, which has been known for some hundreds of year!!!
I bet there is other ways to solve it!!
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