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 $\lim _{x \rightarrow 4} \frac{2 x^2+(3+2 a) x+3 a}{x^3-2 x^2-23 x+60}=\frac{11}{9}$,then $\lim _{x \rightarrow a} \frac{x^2+9 x+20}{x^2-x-20}=$

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:

If $\lim _{x \rightarrow 3} \left( \frac{x^n - 3^n}{x - 3} \right) = 108$ and $n \in N$,then the value of $n$ is

If $\lim _{x \rightarrow 0} \frac{\cos (2 x)+a \cos (4 x)-b}{x^4}$ is finite,then $(a+b)$ is equal to :

If $\mathop {\lim }\limits_{x \to \infty } \left[ {\frac{{{x^3} + 1}}{{{x^2} + 1}} - (ax + b)} \right] = 2$,then