Suppose $$f(x)$$ is differentiable at x = 1 and

$$\mathop {\lim }\limits_{h \to 0} {1 \over h}f\left( {1 + h} \right) = 5$$, then $$f'\left( 1 \right)$$ equals

Let $$\alpha$$ and $$\beta$$ be the distinct roots of $$a{x^2} + bx + c = 0$$, then

$$\mathop {\lim }\limits_{x \to \alpha } {{1 - \cos \left( {a{x^2} + bx + c} \right)} \over {{{\left( {x - \alpha } \right)}^2}}}$$ is equal to

If $$\mathop {\lim }\limits_{x \to \infty } {\left( {1 + {a \over x} + {b \over {{x^2}}}} \right)^{2x}} = {e^2}$$, then the value of $$a$$ and $$b$$, are

Let $$f(x) = {{1 - \tan x} \over {4x - \pi }}$$, $$x \ne {\pi \over 4}$$, $$x \in \left[ {0,{\pi \over 2}} \right]$$.

If $$f(x)$$ is continuous in $$\left[ {0,{\pi \over 2}} \right]$$, then $$f\left( {{\pi \over 4}} \right)$$ is