Define $f(x) = \begin{cases} b - ax & \text{if } x < 2 \\ 3 & \text{if } x = 2 \\ a + 2bx & \text{if } x > 2 \end{cases}$. If $\lim_{x \rightarrow 2} f(x)$ exists,then find the value of $\frac{a}{b}$.

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

If $f(x) = \begin{cases} 4x-5, & x \leq 2 \\ x-k, & x > 2 \end{cases}$ then the value of $k$ for which $\lim_{x \rightarrow 2} f(x)$ exists is equal to:

If the function $f(x)$ satisfies $\lim_{x \rightarrow 1} \frac{f(x)-2}{x^{2}-1} = \pi$,then $\lim_{x \rightarrow 1} f(x) = $

Let $\alpha$ and $\beta$ be the roots of $ax^2 + bx + c = 0$,then $\lim_{x \to \alpha} \frac{1 - \cos(ax^2 + bx + c)}{(x - \alpha)^2}$ is equal to

If $\alpha > \beta > 0$ are the roots of the equation $ax^2 + bx + 1 = 0$,and $\lim_{x}$ ${\rightarrow \frac{1}{\alpha}} \left( \frac{1 - \cos(x^2 + bx + a)}{2(1 - \alpha x)^2} \right)^{\frac{1}{2}} = \frac{1}{k} \left( \frac{1}{\beta} - \frac{1}{\alpha} \right)$,then $k$ is equal to