• # question_answer Let $f(x):\frac{\tan \,[{{e}^{2}}]{{x}^{3}}-\tan \,[-\,{{e}^{2}}]{{x}^{3}}}{{{\sin }^{3}}x},x\ne 0$ ([.] represents greatest integer function). The value of $f(0)$ for which $f(x)$ is continuous, is A) 15                     B) 12        C) $-\,12$                          D) 14

[a]  We have, $f(x)=\frac{\tan [{{e}^{2}}]{{x}^{3}}-\tan [-{{e}^{2}}]{{x}^{3}}}{{{\sin }^{3}}x}$ $7<{{e}^{2}}<8,$ so $[{{e}^{2}}]=7$ and $[-\,{{e}^{2}}]=-\,8$ So $f(0)=\underset{x\to 0}{\mathop{\lim }}\,\frac{\tan 7{{x}^{3}}-\tan (-\,8){{x}^{3}}}{{{\sin }^{3}}x}$ $\underset{x\to 0}{\mathop{\lim }}\,\left[ \frac{\tan 7{{x}^{3}}}{{{\sin }^{3}}x}+\frac{\tan 8{{x}^{3}}}{{{\sin }^{3}}x} \right]$$=7+8=15$