$\lim _{x \rightarrow \frac{\pi}{4}} \frac{2 \sqrt{2}-(\cos x+\sin x)^3}{1-\sin 2 x}=$

Let $[x]$ denote the greatest integer less than or equal to $x$ and $f(x) = 2x - [2x]$. If $\lim_{x \rightarrow 2^{-}} f(x) = l_1$ and $\lim_{x \rightarrow 2^{+}} f(x) = l_2$,then $l_1 + l_2 =$

Let $f(x) = \frac{\sqrt{x+3}}{x+1}$. Then the value of $\lim_{x \rightarrow -3^{-}} f(x)$ is

Let $l = \mathop {\lim}\limits_{x \to 0} \frac{[x]^2}{x^2}$ and $m = \mathop {\lim}\limits_{x \to 0} \frac{[x^2]}{x^2}$,where $[ \cdot ]$ denotes the greatest integer function. Then:

Let $f(x)=5-|x-2|$ and $g(x)=|x+1|, x \in R$. If $f(x)$ attains its maximum value at $\alpha$ and $g(x)$ attains its minimum value at $\beta$,then $\lim _{x \rightarrow-\alpha \beta} \frac{(x-1)\left(x^2-5 x+6\right)}{x^2-6 x+8}$ is equal to

$\lim _{x \rightarrow 0} \frac{63^x-9^x-7^x+1}{\sqrt{2}-\sqrt{1+\cos x}}=\ldots$.