A) 0
B) -3
C) \[-1\]
D) infinity
Correct Answer: B
Solution :
\[\underset{x\to 0}{\mathop{\lim }}\,\frac{{{e}^{x}}+\log (1+x)-{{(1-x)}^{-2}}}{{{x}^{2}}}\] \[\left( \frac{0}{0}form \right)\] By LHospitas rule \[\underset{x\to 0}{\mathop{\lim }}\,\frac{{{e}^{x}}+{{(1-x)}^{-1}}-2{{(1-x)}^{-3}}}{2x}\] \[\left( \frac{0}{0}form \right)\] \[=\underset{x\to 0}{\mathop{\lim }}\,\frac{{{e}^{x}}+{{(1-x)}^{-2}}-6{{(1-x)}^{-4}}}{2}\] (by L Hospitals rule) \[=\frac{{{e}^{0}}-1-6}{2}=\frac{1-1-6}{2}=-3\] Alternative Method \[\underset{x\to 0}{\mathop{\lim }}\,\frac{{{e}^{x}}+\log (1+x)-{{(1-x)}^{-2}}}{{{x}^{2}}}\] \[=\underset{x\to 0}{\mathop{\lim }}\,\frac{\left[ \begin{align} & \left[ 1+x+\frac{{{x}^{2}}}{2!}+\frac{{{x}^{3}}}{3!}+.... \right] \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+\left[ x-\frac{{{x}^{2}}}{2}+\frac{{{x}^{3}}}{3}-\frac{{{x}^{4}}}{4}+.... \right] \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-[1+2x+3{{x}^{2}}+.....] \\ \end{align} \right]}{{{x}^{2}}}\] \[=\underset{x\to 0}{\mathop{\lim }}\,\frac{\left[ -3{{x}^{2}}+-\frac{7}{2}{{(x)}^{3}} \right]}{{{x}^{2}}}\] \[=\underset{x\to 0}{\mathop{\lim }}\,[-3+higher\text{ }powers\text{ }of\,x]\] \[=-3+0=-3\]
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