If $\lim\limits_{x \to \infty} \dfrac{(2x-1)^{19} \cdot (3x+2)^{11}}{(6x-5)^{30}} = 2^a \cdot 3^b$, then $a + b =$

The value of f(0) so that the function $f(x) = \dfrac{(256 - 8x)^{\frac{1}{4}} - 4}{16 - 4(64 + 3x)^{\frac{1}{3}}}$, $x \neq 0$ is continuous at $x = 0$, is

If $\lim\limits_{x \to 0} \dfrac{(4^x - 1)^3}{\tan\left(\dfrac{x}{4}\right)\log\left(1 + \dfrac{x^2}{3}\right)} = 96(\log a)^b$, then $(a + b) = $

If the function f is continuous at $x = 1$, where $f(x) = \dfrac{1 + \cos(\pi x)}{\pi(1-x)^2}$, for $x \neq 1$, then the value of $f(1)$ is....