I have a definition lim theta approaching 0 sin theta/theta = 1

Question

so show if the limit x approaching 0 sin2x/5=2/5

Answer 1

well the point is that sin x ---> x -x^2/2 + .......
as x------> 0
sin x ---> x
so
sin(2x) / 5x ----> [2x - 4x^2 ...]/5x
as x---> 0
sin(2x) /5 ---> 2x/5x = 2/5

Answer 2

you know that

lim(u->0) sinu/u = 1

let u = 2/5 x
lim sin2x/5x
= lim sinu/(5/2 u)
= lim sinu/u * 2/5
= 1 * 2/5
= 2/5