The series for sin(x)=x-x3/3!+x5/5!-...
Also, the series for cos(x)=1-x2/2!+x4/4!-...
sin(x-δ)=sin(x)cos(δ)-cos(x)sin(δ) (trig identity).
If δ is very small, sin(δ)~δ because δ3/3! is negligible. Also, cos(δ)~1 because δ2/2! is negligible.
So sin(x-δ)~sin(x)-δcos(x).
When x=π/2, sin(π/2-δ)~sin(π/2)-δcos(π/2).
Now let x=π/2-δ (δ can be positive or negative).
sin(π/2)=1 and cos(π/2)=0, so sin(π/2-δ)~1 and when δ→0, x→π/2, so in the limit sin(x)→1.