If $\mathop {\lim }\limits_{x \to 0} \frac{{[(a - n)nx - \tan x]\sin nx}}{{{x^2}}} = 0,$ where $n$ is a non-zero real number,then $a$ is equal to

Let $f(x) = \sqrt{\frac{x}{1-x}} + \sqrt{\frac{1-x}{x}}$. If $\lim_{x \rightarrow m} f(x) = 5/2$,then the set of all possible finite values of $m$ is:

Let $f : R - \{0\} \rightarrow R$ be a function such that $f(x) - 6f\left(\frac{1}{x}\right) = \frac{35}{3x} - \frac{5}{2}$. If $\lim_{x \rightarrow 0} \left(\frac{1}{\alpha x} + f(x)\right) = \beta$,where $\alpha, \beta \in R$,then $\alpha + 2\beta$ is equal to:

If $f(x) = \begin{cases} 1+\frac{2x}{a}, & 0 \leq x \leq 1 \\ ax, & 1 < x \leq 2 \end{cases}$,and $\lim_{x \rightarrow 1} f(x)$ exists,then the sum of the cubes of the possible values of $a$ is:

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

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