$$\mathop {\lim }\limits_{x \to 0} {{\sin (\pi {{\sin }^2}x)} \over {{x^2}}} = $$

If the function $$f(x) = \left\{ {\matrix{ {{{{x^2} - (A + 2)x + A} \over {x - 2}},} & {for\,x \ne 2} \cr {2,} & {for\,x = 2} \cr } } \right.$$ is continuous at x = 2, then

$$f(x) = \left\{ {\matrix{ {[x] + [ - x],} & {when\,x \ne 2} \cr {\lambda ,} & {when\,x = 0} \cr } } \right.$$

If f(x) is continuous at x = 2, the value of $$\lambda$$ will be

For the function $$f(x) = {e^{\cos x}}$$, Rolle's theorem is