(D) Let $y=\lim _{x \rightarrow 0} \frac{x e ^x-\log (1+x)}{x^2}$ Applying L-Hospital's rule, we get $y=\lim _{x \rightarrow 0} \frac{ e ^x+x e ^x-\frac{1}{1+x}}{2 x}$ $=\lim _{x \rightarrow 0} \frac{1}{2}\left[ e ^x+ e ^x+x e ^x+\frac{1}{(1+x)^2}\right]$ $=\frac{1}{2}(1+1+0+1)=\frac{3}{2}$
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