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

If $a_1 = 1$ and $a_{n+1} = \frac{4 + 3a_n}{3 + 2a_n}$ for $n \ge 1$,and if $\lim_{n \to \infty} a_n = a$,then the value of $a$ 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

If $\lim _{x \rightarrow \infty}\left(\frac{x^2+x+1}{x+1}-a x-b\right)=4$,then:

If $\mathop {\lim }\limits_{x \to 1} \frac{{{x^4} - 1}}{{x - 1}} = \mathop {\lim }\limits_{x \to k} \frac{{{x^3} - {k^3}}}{{{x^2} - {k^2}}}$,then $k$ is

If $\alpha$ is the positive root of the equation $p(x) = x^{2} - x - 2 = 0$,then $\lim_{x \rightarrow \alpha^{+}} \frac{\sqrt{1 - \cos(p(x))}}{x + \alpha - 4}$ is equal to