If the function $f(x) = (1 - x)\tan \dfrac{{\pi x}}{2}$ is continuous at $x = 1$ ,then $f(1)=$

$\displaystyle\lim_{x\rightarrow 0}\left(\dfrac{1}{\sin^2x}-\dfrac{1}{\sin h^2x}\right)=?$

$\displaystyle \lim_{x \rightarrow 0}\dfrac {1}{x\sqrt {x}}\left(a\ arc\ tan \dfrac {\sqrt {x}}{a}-b\ arc\ \tan \dfrac {\sqrt {x}}{b}\right)$ has the value equal to

Evaluate: $\underset { { x\rightarrow}\dfrac{ \pi }{ 4 } }{ lim } \dfrac { { cot }^{ 3 }x-tanx }{ cos\left( x+\pi /4 \right) } \quad$