If $\mathop {\lim }\limits_{x \to \infty } \left\{ {\ln \left( {{x^2} + 5x} \right) - 2\ln \left( {cx + 1} \right)} \right\} = -2$,then:

Define $f(x) = \begin{cases} \frac{\sqrt{1+px} - \sqrt{1-px}}{x}, & \text{if } -1 \leq x < 0 \\ \frac{2x+1}{x-2}, & \text{if } 0 \leq x \leq 1 \end{cases}$. If $\lim_{x \rightarrow 0} f(x)$ exists,then $p =$

Let $a > 0$ be a root of the equation $2x^2 + x - 2 = 0$. If $\lim_{x \rightarrow \frac{1}{a}} \frac{16(1 - \cos(2 + x - 2x^2))}{1 - ax^2} = \alpha + \beta \sqrt{17}$,where $\alpha, \beta \in \mathbb{Z}$,then $\alpha + \beta$ is equal to:

If $\mathop {\lim }\limits_{x \to 1} \frac{{{x^2} - ax + b}}{{x - 1}} = 3$,then $a + b$ is equal to

$[x]$ represents the greatest integer function. If $\lim _{x \rightarrow 0^{+}} \frac{\cos [x]-\cos (k x-[x])}{x^2}=5$,then $k=$

If the function $f(x)$ satisfies $\mathop {\lim }\limits_{x \to 1} \frac{f(x)-2}{x^{2}-1}=\pi,$ evaluate $\mathop {\lim }\limits_{x \to 1} f(x).$