What is $a$ in $\lim_{t \to 0} \left(\frac{a}{t^2} - \frac{\sin 6t}{t^3 \cos^2 3t}\right) = -18$ 
What is $a$ in $\lim_{t \to 0} \left(\frac{a}{t^2} - \frac{\sin 6t}{t^3 \cos^2 3t}\right) = -18$

We have to make this function defined, by change the cosine into sine or tan. So,
$$\lim_{t \to 0} \left(\frac{a}{t^2} - \frac{2\sin 3t \cos 3t}{t^3 \cos^2 3t}\right) = \lim_{t \to 0} \left(\frac{a}{t^2} - \frac{6}{t^2}\right) = -18$$
But I am confused what should I do next.

Your condition implies that $$\lim_{t\to 0}\frac{at\cos^23t-\sin 6t}{t^3}=-18$$ Adding and subtracting $at$ in numerator we get $$\lim_{t\to 0} \frac{at-\sin 6t}{t^3}-9a\cdot\frac{\sin^23t}{(3t)^2}=-18$$ or $$\lim_{t\to 0}\frac {a-6}{t^2}+\frac{6t-\sin 6t}{t^3}=9(a-2)$$ Using substitution $x=6t$ the second term on left side can be written as $$6^3\lim_{x\to 0}\frac {x-\sin x} {x^3}=36$$ (via L'Hospital's Rule or Taylor series) and therefore we have $$\lim_{t\to 0}\frac {a-6}{t^2}=9(a-6)$$ which implies $a=6$ and the above equation also holds true for this value of $a$. 
