Answer
$$
\lim _{x \rightarrow \infty}\left[x-x^{2} \ln \left(\frac{1+x}{x}\right)\right]=\frac{1}{2}.
$$
Work Step by Step
$$
\begin{aligned}
L& = \lim _{x \rightarrow \infty}\left[x-x^{2} \ln \left(\frac{1+x}{x}\right)\right] \\
&=\lim _{x \rightarrow \infty}\left[x-x^{2} \ln \left(\frac{1}{x}+1\right)\right] \\
& \quad \quad \quad \left[ \text { Let } t=1 / x, \text { so as } x \rightarrow \infty, t \rightarrow 0^{+} , \text { then}\right] \\
&=\lim _{t \rightarrow 0^{+}}\left[\frac{1}{t}-\frac{1}{t^{2}} \ln (t+1)\right]\\
&=\lim _{t \rightarrow 0^{+}} \frac{t-\ln (t+1)}{t^{2}} \\
&
\quad \quad \left[ \text { notice that as } t \rightarrow 0^{+} \text { we have} \quad (t-\ln (t+1)) \rightarrow 0 \quad \text {and} \quad t^{2} \rightarrow 0. \quad \text {This gives an indeterminate form of the type } \frac{0}{0} \\\text {so l’Hospital’s Rule gives} \right] \\
&\stackrel{\mathrm{H}}{=} \lim _{t \rightarrow 0^{+}} \frac{1-\frac{1}{t+1}}{2 t} \\
&=\lim _{t \rightarrow 0^{+}} \frac{t /(t+1)}{2 t} \\
&=\lim _{t \rightarrow 0^{+}} \frac{1}{2(t+1)} \\
&=\frac{1}{2}.
\end{aligned}
$$
