Could you please prove that #lim_(x rarr 0) sin(x)/x = 1# using the formal definition of limits?

1 Answer
Apr 15, 2017

#forall epsilon>0, exists delta=sqrt(6epsilon)>0# s.t. #0 < |x| < delta Rightarrow|sin x/x-1| < epsilon#

Explanation:

Recall:

  1. Error Bound for Alternating Series
    Let #s=sum_(n=0)^(infty)(-1)^nb_n#, where #b_n>0# and #b_n# is monotonically decreasing, and let
    #s_n=sum_(i=0)^n(-1)^n b_n#.
    The error for approximating the sum #s# using the partial sum #s_n# is bounded by #b_(n+1)#, that is, #|s-s_n| leq b_(n+1)#

  2. Power Series of sin x
    #sin x =x-x^3/(3!)+x^5/(5!)-x^7/(7!)+cdots#

So, the error bound for estimating #sin x# by the first term #x# can be expressed as:
#|sin x -x| leq |x^3/(3!)|#

Let us now prove the limit.

#forall epsilon>0, exists delta=sqrt(6epsilon)>0# s.t. #0<|x-0|< delta Rightarrow |x| < sqrt(6epsilon)#

By Error Bound for Alternating Seires,

#Rightarrow |sin x/x -1|=|(sin x - x)/x|leq|x^3/(3!)|/|x|=x^2/6<(sqrt(6epsilon))^2/6=epsilon#

Hence, #lim_(x to 0)sinx/x=1#