When x→0, arcsinx ~ x, so arcsin2x ~ 2x.

lim { x→0 } arcsin 2x/(5x)= lim { x→0 } 2x/(5x)= 2/5

2. Defined by derivative, original limit = sin' x = cos x

3 x[ln(x+ 1)-ln x]= x ln( 1+ 1/x)

When x → 0, ln( 1+x) ~ x

Therefore, when x → +∞, ln (1+1/x) ~1/x.

So the original limit = lim {x→ +∞} x *1/x =1.