The value of $\mathop {\lim }\limits_{x \to 0} \left( \frac{e^x - 1}{x} \right)$ is

If $\lim_{n \rightarrow \infty} \frac{(n+1)^{k-1}}{n^{k+1}}[(nk+1)+(nk+2)+\ldots+(nk+n)] = 33 \cdot \lim_{n \rightarrow \infty} \frac{1}{n^{k+1}} \cdot [1^k + 2^k + 3^k + \ldots + n^k]$,then the integral value of $k$ is equal to $....$

Let $[P]$ denote the greatest integer $\leq P$. If $0 \leq a \leq 2$,then the number of integral values of $a$ such that $\lim _{x \rightarrow a}([x^2]-[x]^2)$ does not exist is:

Let $[x]$ denote the greatest integer not exceeding $x$. If $l_1 = \lim_{x \rightarrow 2^{+}} (x^2 + [x])$,$l_2 = \lim_{x \rightarrow 3^{-}} (2x - [x])$ and $l_3 = \lim_{x \rightarrow \frac{\pi}{2}} \left( \frac{\cos x}{x - \frac{\pi}{2}} \right)$,then:

If $a, b, c, d$ are positive,then $\mathop {\lim }\limits_{x \to \infty } {\left( {1 + \frac{1}{{a + bx}}} \right)^{c + dx}} = $

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