If $g(x) = \frac{x}{[x]}$ for $x > 2$,then $\lim_{x \rightarrow 2^+} \frac{g(x) - g(2)}{x - 2}$ is equal to

$\mathop {\lim }\limits_{x \to \infty } \frac{{\sqrt {{x^2} - \sqrt {{x^2} - \sqrt {{x^2} - \dots} } } }}{x}$ is equal to-

Find $\mathop {\lim }\limits_{x \to 0} f(x)$ and $\mathop {\lim }\limits_{x \to 1} f(x),$ where $f(x) = \begin{cases} 2x+3, & x \leq 0 \\ 3(x+1), & x > 0 \end{cases}$

If $S_1 = \sum_{r=1}^{n} r$,$S_2 = \sum_{r=1}^{n} r^2$,and $S_3 = \sum_{r=1}^{n} r^3$,then the value of $\lim_{n \rightarrow \infty} \frac{S_1(1 + \frac{S_3}{4})}{S_2^2}$ is

Let for all $x > 0$, $f(x) = \lim_{n \rightarrow \infty} n(x^{1/n} - 1)$, then

Let $e$ denote the base of the natural logarithm. The value of the real number $a$ for which the right-hand limit $\lim_{x \rightarrow 0^{+}} \frac{(1-x)^{\frac{1}{x}}-e^{-1}}{x^a}$ is equal to a nonzero real number is: