The function $$f:R/\left\{ 0 \right\} \to R$$ given by

$$f\left( x \right) = {1 \over x} - {2 \over {{e^{2x}} - 1}}$$

can be made continuous at $$x$$ = 0 by defining $$f$$(0) as

The set of points where $$f\left( x \right) = {x \over {1 + \left| x \right|}}$$ is differentiable is

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

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