If $\mathop {\lim }\limits_{x \to \frac{1}{2}} \frac{{a{x^2} + bx + c}}{{{{(2x - 1)}^2}}} = \frac{1}{2}$,then $\mathop {\lim }\limits_{x \to 2} \frac{{(x - a)(x - b)(x - c)}}{{x - 2}}$ is

If $f(x) = \begin{cases} 3ax - 2b, & x > 1 \\ ax + b + 1, & x < 1 \end{cases}$ and $\lim_{x \rightarrow 1} f(x)$ exists,then the relation between $a$ and $b$ is

If $\lim _{x}$ ${\rightarrow \infty} \frac{(\sqrt{2 x+1}+\sqrt{2 x-1})^8+(\sqrt{2 x+1}-\sqrt{2 x-1})^8(P x^4-16)}{(x+\sqrt{x^2-2})^8+(x-\sqrt{x^2-2})^8} = 1$,then $P=$

Let $\alpha(a)$ and $\beta(a)$ be the roots of the equation $(\sqrt[3]{1+a}-1) x^2+(\sqrt{1+a}-1) x+(\sqrt[6]{1+a}-1)=0$ where $a > -1$. Then $\lim _{a \rightarrow 0^{+}} \alpha(a)$ and $\lim _{a \rightarrow 0^{+}} \beta(a)$ are

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 $\frac{a}{b}$.

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 $p =$