A) 1
B) 0
C) 1/2
D) \[-1\]
E) 2
Correct Answer: A
Solution :
\[f(x)=\frac{{{\log }_{e}}(1+{{x}^{2}}\tan x)}{\sin {{x}^{3}}}\] This function is continuous at\[x=0,\]then \[\underset{x\to 0}{\mathop{\lim }}\,\frac{{{\log }_{e}}(1+{{x}^{2}}\tan x)}{\sin {{x}^{3}}}=f(0)\] \[\Rightarrow \]\[\underset{x\to 0}{\mathop{\lim }}\,\frac{{{\log }_{e}}\left\{ 1+{{x}^{2}}\left( x+\frac{{{x}^{3}}}{3}+\frac{2{{x}^{5}}}{15}+..... \right) \right\}}{{{x}^{3}}-\frac{{{x}^{9}}}{3!}+\frac{{{x}^{15}}}{5!}-....}\] \[\Rightarrow \]\[\underset{x\to 0}{\mathop{\lim }}\,\frac{{{\log }_{e}}(1+{{x}^{3}})}{{{x}^{3}}-\frac{{{x}^{9}}}{3!}+\frac{{{x}^{15}}}{5!}-....}=f(0)\] [on neglecting higher power of\[x\]in\[{{x}^{2}}tan\text{ }x\]] \[\Rightarrow \]\[\underset{x\to 0}{\mathop{\lim }}\,\frac{{{x}^{3}}-\frac{{{x}^{6}}}{2}+\frac{{{x}^{9}}}{3}-.....}{{{x}^{3}}-\frac{{{x}^{9}}}{3!}+\frac{{{x}^{15}}}{5!}-....}=f(0)\] \[\Rightarrow \] \[1=f(0)\]
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