Find a so that limit is finite

Let L = lim_{x→0} (a sin x − sin 2x)/tan³x. The denominator tan³x → 0 like x³, so for the limit to be finite the numerator must also vanish to the same order. Use series:

a sin x   = a(x − x³/6 + ...)
sin 2x    = 2x − (2x)³/6 + ... = 2x − 4x³/3 + ...
numerator = (a − 2)x + (−a/6 + 4/3)x³ + ...
tan³x     = x³ + ...

Dividing by x³, the term (a − 2)/x² blows up unless a − 2 = 0. So a = 2.

With a = 2 the x³ coefficient of the numerator is −2/6 + 4/3 = −1/3 + 4/3 = 1, so

L = lim (x³ + ...)/(x³ + ...) = 1.

(By L'Hospital's rule: the numerator derivative a cos x − 2 cos 2x must vanish at 0, giving a = 2. This confirms the value.)

Answer: a = 2 and the limit is 1.