If the function $$f(x) = \left\{ {\matrix{ {{{{{\log }_e}(1 - x + {x^2}) + {{\log }_e}(1 + x + {x^2})} \over {\sec x - \cos x}}} & , & {x \in \left( {{{ - \pi } \over 2},{\pi \over 2}} \right) - \{ 0\} } \cr k & , & {x = 0} \cr } } \right.$$ is continuous at x = 0, then k is equal to:

<p>$$f(x) = \left\{ {\matrix{ {{{{{\log }_e}(1 - x + {x^2}) + {{\log }_e}(1 + x + {x^2})} \over {\sec x - \cos x}}} & , & {x \in \left( {{{ - \pi } \over 2},{\pi \over 2}} \right) - \{ 0\} } \cr k & , & {x = 0} \cr } } \right.$$</p> <p>for continuity at $x = 0$</p> <p>$\mathop {\lim }\limits_{x \to 0} f(x) = k$</p> <p>$\therefore$ $$k = \mathop {\lim }\limits_{x \to 0} {{{{\log }_e}({x^4} + {x^2} + 1)} \over {\sec x - \cos x}}\left( {{0 \over 0}\,\mathrm{form}} \right)$$</p> <p>$$ = \mathop {\lim }\limits_{x \to 0} {{\cos x{{\log }_e}({x^4} + {x^2} + 1)} \over {{{\sin }^2}x}}$$</p> <p>$$ = \mathop {\lim }\limits_{x \to 0} {{{{\log }_e}({x^4} + {x^2} + 1)} \over {{x^2}}}$$</p> <p>$$ = \mathop {\lim }\limits_{x \to 0} {{\ln (1 + {x^2} + {x^4})} \over {{x^2} + {x^4}}}\,.\,{{{x^2} + {x^4}} \over {{x^2}}}$$</p> <p>$= 1$</p>