How do you find the limit of #(x+1-cos(x))/(4x)# as x approaches 0?

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Answer 1 · 2016-05-30T12:32:53

#lim_{x->0}(x + 1 - Cos(x))/(4 x) = 1/4#

We will be using

#Cos(a+b)=Cos(a)Cos(b) - Sin(a) Sin(b)#

and

#sin^2(a) + cos^2(a) = 1#

Using #cos(x) = 1-2 sin^2(x/2)#

and substituting

#(x + 1 - Cos(x))/(4 x) = (x +2sin^2 (x/2))/(4x)#
#lim_{x->0}(x +2sin^2 (x/2))/(4x)=lim_{x->0}(x/(4x))+lim_{x->0}(2 sin^2 (x/2))/(4x)#

but

#lim_{x->0}(2 sin^2 (x/2))/(4x) = lim_{x->0}sin(x/2)/4lim_{x->0}(sin(x/2)/(x/2)) = 0 times 1 #

so

#lim_{x->0}(x + 1 - Cos(x))/(4 x) = 1/4#