Question

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

Answer 1

`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`