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

If $\lim _{x \rightarrow 0}\left(\frac{\cos 4 x+a \cos 2 x+b}{x^4}\right)$ is finite,then the values of $a, b$ are respectively :

If $\lim _{n \rightarrow \infty}\left(\sqrt{n^{2}-n-1}+n \alpha+\beta\right)=0$,then $8(\alpha+\beta)$ is equal to:

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

The values of the constants $\alpha$ and $\beta$ such that $\lim_{x \to \infty} \left( \frac{x^2 + 1}{x + 1} - \alpha x - \beta \right) = 0$ are respectively:

The product of all possible values of $\alpha$, for which $\lim_{x \to 0} \left( \frac{1 - \cos(\alpha x) \cos((\alpha + 1)x) \cos((\alpha + 2)x)}{\sin^2((\alpha + 1)x)} \right) = 2$, is: