Let $\alpha$ $\in$ R be such that the function $$f(x) = \left\{ {\matrix{ {{{{{\cos }^{ - 1}}(1 - {{\{ x\} }^2}){{\sin }^{ - 1}}(1 - \{ x\} )} \over {\{ x\} - {{\{ x\} }^3}}},} & {x \ne 0} \cr {\alpha ,} & {x = 0} \cr } } \right.$$ is continuous at x = 0, where {x} = x $-$ [ x ] is the greatest integer less than or equal to x. Then :

$$RHL = \mathop {\lim }\limits_{x \to {0^ + }} {{{{\cos }^{ - 1}}(1 - {x^2}){{\sin }^{ - 1}}(1 - x)} \over {x(1 - {x^2})}} $$ <br><br>$$= {\pi \over 2}\mathop {\lim }\limits_{x \to {0^ + }} {{{{\cos }^{ - 1}}(1 - {x^2})} \over x}$$<br><br>$$ = {\pi \over 2}\mathop {\lim }\limits_{x \to {0^ + }} {{ - 1} \over {\sqrt {1 - {{(1 - {x^2})}^2}} }}( - 2x)$$ (L' Hospital Rule)<br><br>$$ = \pi \mathop {\lim }\limits_{x \to {0^ + }} {x \over {\sqrt {2{x^2} - {x^4}} }} = \pi \mathop {\lim }\limits_{x \to {0^ + }} {1 \over {\sqrt {2 - {x^2}} }} = {\pi \over {\sqrt 2 }}$$<br><br>$$LHL = \mathop {\lim }\limits_{x \to {0^ - }} {{{{\cos }^{ - 1}}(1 - {{(1 + x)}^2}){{\sin }^{ - 1}}( - x)} \over {(1 + x) - {{(1 + x)}^3}}} $$ <br><br>$$= {\pi \over 2}\mathop {\lim }\limits_{x \to {0^ - }} {{{{\sin }^{ - 1}}x} \over {(1 - x)\left[ {{{(1 + x)}^2} - 1} \right]}} = {\pi \over 2}\mathop {\lim }\limits_{x \to {0^ - }} {{{{\sin }^{ - 1}}x} \over {{x^2} + 2x}}$$<br><br>$= {\pi \over 2}\left( {{1 \over 2}} \right) = {\pi \over 4}$<br><br>As LHL $\ne$ RHL so f(x) is not continuous at x = 0