mathop {lim }limits_{x to 0} frac{{x{e^x} - l | vedclass

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(D) Let $y = \mathop {\lim }\limits_{x 	o 0} \frac{{x{e^x} - \log (1 + x)}}{{{x^2}}}$.Since this is a $\frac{0}{0}$ form,we apply $L'	ext{Hospital's

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$\frac{3}{2}$

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(D) Let $y = \mathop {\lim }\limits_{x 	o 0} \frac{{x{e^x} - \log (1 + x)}}{{{x^2}}}$.Since this is a $\frac{0}{0}$ form,we apply $L'	ext{Hospital's rule}$.$y = \mathop {\lim }\limits_{x 	o 0} \frac{{\frac{d}{{dx}}(x{e^x} - \log (1 + x))}}{{\frac{d}{{dx}}({x^2})}}$$y = \mathop {\lim }\limits_{x 	o 0} \frac{{{e^x} + x{e^x} - \frac{1}{{1 + x}}}}{{2x}}$.Again,this is a $\frac{0}{0}$ form,so we apply $L'	ext{Hospital's rule}$ again.$y = \mathop {\lim }\limits_{x 	o 0} \frac{{\frac{d}{{dx}}({e^x} + x{e^x} - \frac{1}{{1 + x}})}}{{\frac{d}{{dx}}(2x)}}$$y = \mathop {\lim }\limits_{x 	o 0} \frac{{{e^x} + {e^x} + x{e^x} + \frac{1}{{{{(1 + x)}^2}}}}}{2}$Substituting $x = 0$:$y = \frac{{{e^0} + {e^0} + 0 \cdot {e^0} + \frac{1}{{{{(1 + 0)}^2}}}}}{2} = \frac{{1 + 1 + 0 + 1}}{2} = \frac{3}{2}$.

$\mathop {\lim }\limits_{x \to 0} \frac{{x{e^x} - \log (1 + x)}}{{{x^2}}}$ equals

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