If $[ \cdot ]$ denotes the greatest integer function,then evaluate the limit: $\lim _{x \rightarrow \frac{\pi^{+}}{2}} \frac{[\sin x]-[\cos x]+1}{2}$

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

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

$\lim _{x \rightarrow 0} \left( \frac{\tan x}{\sqrt{2x+4}-2} \right)$ is equal to

If $\mathop {\lim }\limits_{x \to 0} \frac{{\log (a + x) - \log a}}{x} + k\mathop {\lim }\limits_{x \to e} \frac{{\log x - 1}}{{x - e}} = 1$,then

$\lim _{n \rightarrow \infty}\left\{n-\sqrt{n^2-4 n}\right\}=$