$$\mathop {\lim }\limits_{x \to {\pi \over 2}} {{\left[ {1 - \tan \left( {{x \over 2}} \right)} \right]\left[ {1 - \sin x} \right]} \over {\left[ {1 + \tan \left( {{x \over 2}} \right)} \right]{{\left[ {\pi - 2x} \right]}^3}}}$$ is

If $$f(x) = \left\{ {\matrix{ {x{e^{ - \left( {{1 \over {\left| x \right|}} + {1 \over x}} \right)}}} & {,x \ne 0} \cr 0 & {,x = 0} \cr } } \right.$$

then $$f(x)$$ is

$$\mathop {\lim }\limits_{x \to 0} {{\sqrt {1 - \cos 2x} } \over {\sqrt 2 x}}$$ is

Let $$f(2) = 4$$ and $$f'(x) = 4.$$

Then $$\mathop {\lim }\limits_{x \to 2} {{xf\left( 2 \right) - 2f\left( x \right)} \over {x - 2}}$$ is given by