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

Let $\tan (2\pi |\sin \theta |) = \cot (2\pi |\cos \theta |)$,where $\theta \in R$ and $f(x) = (\sin^2 \theta + \cos^2 \theta)$. The value of $\lim_{x \to \infty} [\frac{2}{f(x)}]$ equals (Here $[\,]$ represents the greatest integer function).

If $\lim _{x \rightarrow 0} \frac{e^{a x}-\cos (b x)-\frac{c x e^{-c x}}{2}}{1-\cos (2 x)}=17$,then $5 a^2+b^2$ is equal to

Given $f(x) = \frac{ax + b}{x + 1}$,$\lim_{x \rightarrow \infty} f(x) = 1$ and $\lim_{x \rightarrow 0} f(x) = 2$,then $f(-2)$ is

If $\lim _{x \rightarrow 0} \frac{\alpha x e^{x}-\beta \log _{e}(1+x)+\gamma x^{2} e^{-x}}{x \sin ^{2} x}=10$,where $\alpha, \beta, \gamma \in R$,then the value of $\alpha+\beta+\gamma$ is:

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