$\displaystyle\lim_{x \to 0}\dfrac{(5^x - 1)^4\,\text{cosec}\,(x\log 5)}{\tan(x\log 5) \cdot \log(1 + x^2\log 25)} = \ldots\ldots$

If the derivative of the function $f(x) = \begin{cases} ax^2 + b & \text{if } x < -1 \\ bx^2 + ax + 4 & \text{if } x \geq -1 \end{cases}$ is continuous everywhere then

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