200 days streak special 6

In order for the above limit to equal unity as $x \rightarrow 0$ , one must apply L'Hopital's Rule thrice in order for the numerator to be non-zero. Let $f(x) = x^3$ and $g(x) =\sqrt{x+a} \cdot (bx - \sin{x})$ with third-derivatives equaling (via Wolfram-Alpha for brevity):

$f'''(x) = 6$ ;

$g'''(x) = \frac{3(bx - \sin{x})}{8(x+a)^{\frac{5}{2}}} - \frac{3(b - \cos{x})}{4(x+a)^{\frac{3}{2}}} + \frac{3\sin{x}}{2\sqrt{x+a}} + \cos{x}\sqrt{x+a}$

which the limit computes to:

$\lim_{x \rightarrow 0} \frac{f'''(x)}{g'''(x)} = \frac{6}{\sqrt{a} - \frac{3(b-1)}{4a\sqrt{a}}} = \frac{24a\sqrt{a}}{4a^2 - 3(b-1)} = 1$

which occurs for $a = 36, b = 1$ . Hence, $a + b + 1979 = \boxed{2016}.$