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 $f: R \to [0, \infty)$ is such that $\lim_{x \to 5} f(x)$ exists and $\lim_{x \to 5} \frac{(f(x))^2 - 9}{\sqrt{|x - 5|}} = 0$,then $\lim_{x \to 5} f(x)$ equals:

If $\lim _{x \rightarrow 1} \frac{x^2-ax+b}{x-1}=7$,then $a+b$ is equal to

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:

Let $\alpha, \beta \in R$ be such that $\lim _{x \rightarrow 0} \frac{x^2 \sin (\beta x)}{\alpha x-\sin x}=1$. Then $6(\alpha+\beta)$ equals

If $\lim _{x \rightarrow 2} \frac{3 x^2-a x+5 b}{x-2}=17$,then $a b=$