Question #8b5f0
1 Answer
Explanation:
What we will need:
#lim_(x->0)sin(x)/x = 1#
This can be derived using the squeeze theorem together with some geometric arguments.
#{(lim_(x->a)f(x) = L_f), (lim_(x->a)g(x) = L_g):}=>lim_(x->a)f(x)g(x) = L_fL_g#
It can be shown using the definition of a limit that if two functions have finite limits at a point, then the limit of the sum of those functions is the sum of their limits at that point.
#{(lim_(x->a)f(x) = L_f), (lim_(x->a)g(x) = L_g):}=>lim_(x->a)(f(x)+g(x)) = L_f+L_g#
A similar property applies to sums.
#lim_(x->0)(cos(x)-1)/x = 0#
This can be derived from the first limit shown above by using some algebraic manipulation to show that
#d/dxf(x) = lim_(h->0)(f(x+h)-f(x))/h#
The definition of a derivative.
With the above, we have
#=lim_(h->0)(sin(x+h)-sin(x))/h#
#=lim_(h->0)(sin(x)cos(h)+cos(x)sin(h)-sin(x))/h#
#=lim_(h->0)(sin(x)(cos(h)-1) + cos(x)sin(h))/h#
#=lim_(h->0)(sin(x)(cos(h)-1)/h + cos(x)sin(h)/h)#
#=lim_(h->0)sin(x)(cos(h)-1)/h + lim_(h->0)cos(x)sin(h)/h#
#=sin(x)*0 + cos(x)*1#
#=cos(x)#