$\mathop {\lim }\limits_{x \to 0} \left( {\frac{{{a^x} - {b^x}}}{x}} \right) = $

If $\lim _{x \rightarrow \infty}\left(1+\frac{p}{x}\right)^{q x}=e^9$ where $p, q \in \mathbb{N}$,then $p+q=$

Let $f(x)=5-|x-2|$ and $g(x)=|x+1|, x \in R$. If $f(x)$ attains its maximum value at $\alpha$ and $g(x)$ attains its minimum value at $\beta$,then $\lim _{x \rightarrow-\alpha \beta} \frac{(x-1)\left(x^2-5 x+6\right)}{x^2-6 x+8}$ is equal to

$\mathop {\lim }\limits_{x \to \infty } {\left[ {1 + \frac{1}{{mx}}} \right]^x}$ is equal to

Assertion $(A)$: $\lim _{x \rightarrow 0} \frac{1}{x} = \infty$
Reason $(R)$: As the value of $x$ decreases,the value of $\frac{1}{x}$ increases.

$\lim _{x \rightarrow 3 / 2} \frac{\left(4 x^2-6 x\right)\left(4 x^2+6 x+9\right)}{\sqrt[3]{2 x}-\sqrt[3]{3}}=$