Using the epsilon-delta definition of a limit, show that limit of sin(x) as x approaches pi/2 is 1.

Question

Please use as few theorems from real analysis as possible (I know how to do a proof using the mean value theorem but was wondering if there is a more basic way to do this for people who are just starting real analysis). Using Taylor series is ok.

Answer 1

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.