The function $$f(x) = {{1 - \sin x + \cos x} \over {1 + \sin x + \cos x}}$$ is not defined at x = $$\pi$$. The value of f($$\pi$$), so that f(x) is continuous at x = $$\pi$$, is

Consider the following functions

1. $$f(x) = \left\{ \matrix{ {1 \over x},\,if\,x \ne 0 \hfill \cr 0,\,if\,x = 0 \hfill \cr} \right.$$

$$f(x) = \left\{ {\matrix{ {2x + 5,} & {if\,x > 0} \cr {{x^2} + 2x + 5,} & {if\,x \le 0} \cr } } \right.$$

Which of the above functions is/are derivable at x = 0 ?

If $$f(x) = {{\sin ({e^{x - 2}} - 1)} \over {\ln (x - 1)}}$$, then $$\mathop {\lim }\limits_{x \to 2} $$ f(x) is equal to

Consider the function

$$f(x) = \left\{ {\matrix{ { - 2\sin x,} & {if} & {x \le - {\pi \over 2}} \cr {A\sin x + B,} & {if} & { - {\pi \over 2} < x < {\pi \over 2}} \cr {\cos x,} & {if} & {x \ge {\pi \over 2}} \cr } } \right.$$

which is continuous everywhere.

The value of A is