The limit $\lim _{x \rightarrow 1} \frac{\sqrt{1-\cos 2(x-1)}}{x-1}$

$\mathop {\lim }\limits_{x \to 1} \frac{{1 - {x^{ - 1/3}}}}{{1 - {x^{ - 2/3}}}} = $

If $\alpha$ and $\beta$ are the roots of the equation $375x^2 - 25x - 2 = 0$,then $\lim_{n \to \infty} \sum_{r=1}^n \alpha^r + \lim_{n \to \infty} \sum_{r=1}^n \beta^r$ is equal to

If $f(x) = \frac{\sin(e^{x-2} - 1)}{\log(x-1)},$ then $\lim_{x \to 2} f(x)$ is given by

$\lim _{x}$ ${\rightarrow \infty} \frac{(2 x+1)^{50}+(2 x+2)^{50}+(2 x+3)^{50}+\cdots \cdots+(2 x+100)^{50}}{(2 x)^{50}+(10)^{50}} = \dots$

Let $[.]$ denote the greatest integer function. Assertion $(A) : \lim_{x \rightarrow \infty} \frac{[x]}{x} = 1$. Reason $(R) : f(x) = x - 1, g(x) = [x], h(x) = x$ and $\lim_{x \rightarrow \infty} \frac{f(x)}{x} = \lim_{x \rightarrow \infty} \frac{h(x)}{x} = 1$.