A) \[\frac{2}{3}\]
B) \[\frac{1}{3}\]
C) \[\frac{1}{2}\]
D) \[\frac{3}{2}\]
Correct Answer: D
Solution :
Let \[y=\underset{x\to 0}{\mathop{\lim }}\,\,\,\frac{x\,{{e}^{x}}-\log \,(1+x)}{{{x}^{2}}}\], \[\left( \frac{0}{0}\,\text{form} \right)\] Applying L-Hospital's rule, \[y=\underset{x\to 0}{\mathop{\lim }}\,\,\,\frac{{{e}^{x}}+x\,{{e}^{x}}-\frac{1}{1+x}}{2x}\], \[\left( \frac{0}{0}\,\text{form} \right)\] \[y=\underset{x\to 0}{\mathop{\lim }}\,\,\,\frac{1}{2}\,\left[ {{e}^{x}}+{{e}^{x}}+x\,{{e}^{x}}+\frac{1}{{{(1+x)}^{2}}} \right]\] \[y=\underset{x\to 0}{\mathop{\lim }}\,\,\,\frac{1}{2}\,[1+1+0+1]=\frac{3}{2}\].
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