If the function $f$ defined on $\left(\frac{\pi}{6}, \frac{\pi}{3}\right)$ by

$$f(x)=\left\{\begin{array}{cc} \frac{\sqrt{2} \cos x-1}{\cot x-1}, & x \neq \frac{\pi}{4} \\ k \quad, & x=\frac{\pi}{4} \end{array}\right.$$

is continuous, then k is equal to

The value of $\lim _\limits{x \rightarrow 0}\left((\sin x)^{\frac{1}{x}}+\left(\frac{1}{x}\right)^{\sin x}\right)$, where $x>0$ is

Let $\mathrm{f}(x)=\frac{1-\tan x}{4 x-\pi}, x \neq \frac{\pi}{4}, x \in\left[0, \frac{\pi}{2}\right]$. $f(x)$ is continuous in $\left[0, \frac{\pi}{2}\right]$, then $f\left(\frac{\pi}{4}\right)$ is

If the function $f(x)= \begin{cases}-2 \sin x & \text {, if } x \leq \frac{-\pi}{2} \\ A \sin x+B & , \text { if } \frac{-\pi}{2}< x<\frac{\pi}{2} \\ \cos x & , \text { if } x \geq \frac{\pi}{2}\end{cases}$ is continuous everywhere, then the values of $A$ and B are respectively