A) 15
B) 12
C) \[-\,12\]
D) 14
Correct Answer: A
Solution :
[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 \[f(0)=\underset{x\to 0}{\mathop{\lim }}\,\frac{\tan 7{{x}^{3}}-\tan (-\,8){{x}^{3}}}{{{\sin }^{3}}x}\] |
\[\underset{x\to 0}{\mathop{\lim }}\,\left[ \frac{\tan 7{{x}^{3}}}{{{\sin }^{3}}x}+\frac{\tan 8{{x}^{3}}}{{{\sin }^{3}}x} \right]\]\[=7+8=15\] |
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