Lim(h→0) (2sin(5h) − 30h)/ (sin(6h))

Calculus help. The answer is not 0

lim [2sin(5h)-30h]/sin(6h)
h->0

lim {2[(sin(5h)/5h)]*x-30h] / [(sin(6h)/6h)*6h]

lim sin(5h)/5h=1
h->0
lim sin(6h)/6h =1
h->0

lim (2*1*5h-30h)/(1*6h)
h->0

lim -20/6
h->0

the limit is -20/6 =-10/3

Make use of
lim[x->0] x/sin(x) = lim[x->0] sin(x)/x = 1

substitutions, the properties of limits, and "fancy multiplication by 1."

Lim(h→0) (2sin(5h) − 30h)/ (sin(6h)) =
Lim(h→0) (2sin(5h) /h − 30)/ (sin(6h)/h) =
Lim(h→0) (10sin(5h) /(5h) − 30)/ (6sin(6h)/(6h)) =( 10 - 30)/6 = -10/3