Evaluate the given limit: $\mathop {\lim }\limits_{x \to 0} \frac{ax + x \cos x}{b \sin x}$

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

Let $f(x) = \frac{x \cdot 2^x - x}{1 - \cos x}$ and $g(x) = 2^x \sin \left( \frac{\ln 2}{2^x} \right)$,then:

$\mathop {\lim }\limits_{x \to 1} f(x)$ is equal to,where $f(x) = \begin{cases} \frac{e^{\frac{1}{x-1}} - 2}{e^{\frac{1}{x-1}} + 2} & x \neq 1 \\ 1 & x = 1 \end{cases}$

The value of $\lim _{x \rightarrow 0^{+}} \frac{\cos ^{-1}\left(x-[x]^{2}\right) \cdot \sin ^{-1}\left(x-[x]^{2}\right)}{x-x^{3}},$ where $[x]$ denotes the greatest integer $\leq x$ is

If $0 < p < q$,then $\lim _{n \rightarrow \infty}\left(q^n+p^n\right)^{1 / n}$ is equal to :