If $$f\left( x \right)$$ is continuous in [0, 1] and $$f\left( {\frac{1}{3}} \right) = 1,$$ then $$\mathop {\lim }\limits_{n \to \infty } f\left( {\frac{n}{{\sqrt {9{n^2} + 1} }}} \right)$$ is equal to :
If $$G\left( x \right) = - \sqrt {25 - {x^2}} $$ then $$\mathop {\lim }\limits_{x\, \to \,{\text{I}}} \frac{{G\left( x \right) - G\left( I \right)}}{{x - 1}}$$ has the value-
If $$\eqalign{
& f\left( x \right) = \frac{{\sin \left[ x \right]}}{{\left[ x \right]}},\,\,\left[ x \right] \ne 0 \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\, = 0,\,\,\,\,\,\,\,\,\,\,\,\,\,\left[ x \right] = 0 \cr} $$
Where \[\left[ x \right]\] denotes the greatest integer less than or equal to $$x.$$ then $$\mathop {\lim }\limits_{x\, \to \,0} f\left( x \right)$$ equals