If $\lim _{x \rightarrow 5} \frac{x^{k}-5^{k}}{x-5}=500$,then the value of $k$,where $k \in N$ is

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

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

If $f(x) = \begin{cases} 4x-5, & x \leq 2 \\ x-k, & x > 2 \end{cases}$ then the value of $k$ for which $\lim_{x \rightarrow 2} f(x)$ exists 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:

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