If $f(x) = \begin{cases} x, & x > 1 \\ x^2, & x < 1 \end{cases}$,then $\lim_{x \to 1} f(x) = $

The function $f(x) = \cot x$ is discontinuous at every point of the set

The function $f(x) = \frac{1 - \sin x + \cos x}{1 + \sin x + \cos x}$ is not defined at $x = \pi$. The value of $f(\pi)$,so that $f(x)$ is continuous at $x = \pi$,is

If $\{x\} = x - [x]$ where $[x]$ is the greatest integer $\leq x$ and $\lim_{x \rightarrow 0^{-}} \frac{\cos^{-1}(1-\{x\}^2) \sin^{-1}(1-\{x\})}{\{x\}-\{x\}^4} = \theta$,then $\tan \theta =$

$f(x) = \begin{cases} \frac{(2x^2 - ax + 1) - (ax^2 + 3bx + 2)}{x + 1} & ; x \neq -1 \\ k & ; x = -1 \end{cases}$ is a real-valued function. If $a, b, k \in R$ and $f$ is continuous on $R$,then $k =$

Let $x=2$ be a root of the equation $x^2+px+q=0$ and $f(x)=\begin{cases} \frac{1-\cos(x^2-4px+q^2+8q+16)}{(x-2p)^4}, & x \neq 2p \\ 0, & x=2p \end{cases}$. Then $\lim _{x \rightarrow 2p^{+}}[f(x)]$,where $[.]$ denotes the greatest integer function,is $........$