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$\lim _{x \rightarrow 0} \frac{x e^x-\log (1+x)}{x^2}$ equals
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Verified Answer
The correct answer is:
$\frac{3}{2}$
$\text { Let } y=\lim _{x \rightarrow 0} \frac{x e^x-\log (1+x)}{x^2},\left(\frac{0}{0} \text { form }\right)$
Applying L-Hospital's rule,
$\begin{aligned}
& y=\lim _{x \rightarrow 0} \frac{e^x+x e^x-\frac{1}{1+x}}{2 x},\left(\frac{0}{0} \text { form }\right) \\
& y=\lim _{x \rightarrow 0} \frac{1}{2}\left[e^x+e^x+x e^x+\frac{1}{(1+x)^2}\right] \\
& y=\lim _{x \rightarrow 0} \frac{1}{2}[1+1+0+1]=\frac{3}{2}
\end{aligned}$
Applying L-Hospital's rule,
$\begin{aligned}
& y=\lim _{x \rightarrow 0} \frac{e^x+x e^x-\frac{1}{1+x}}{2 x},\left(\frac{0}{0} \text { form }\right) \\
& y=\lim _{x \rightarrow 0} \frac{1}{2}\left[e^x+e^x+x e^x+\frac{1}{(1+x)^2}\right] \\
& y=\lim _{x \rightarrow 0} \frac{1}{2}[1+1+0+1]=\frac{3}{2}
\end{aligned}$
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