Compute:
lim (sin3x-sin5x)/(3sin2x) as x approaches 0.
recall that sin(u)/u → 1 as x→0
so now we have
3 sin3x/3x - 5 sin5x/5x → 3-5 = -2
2*3sin2x /2x → 6
so the whole limit is -2/6 = -1/3
all those x's in the denominator cancel out
sin(3x) = 3sinx - 4 sin^3 x
sin(5x) = 16 sin^5 x - 20 sin^3 x + 5 sin x
(sin3x-sin5x)/(3sin2x)
= (3sinx - 4 sin^3 x - (16 sin^5 x - 20 sin^3 x + 5 sin x ))/(6sinx cosx)
= (-16sin^5 x + 16sin^3 x - 2sinx)/(6sinxcosx)
= (-16 sin^4 x + 16sin^2 x - 2)/6cosx , where sinx ≠ 0
limit (-16 sin^4 x + 16sin^2 x - 2)/6cosx as x ---> 0
= (0 + 0 - 2)/6 = -1/3