Answer
$$
\begin{aligned}
L&=\lim _{x \rightarrow \infty}\left[x-x^{2} \ln \left(\frac{1+x}{x}\right)\right]\\
&=\frac{1}{2}
\end{aligned}
$$
Work Step by Step
$$
\begin{aligned}
L&=\lim _{x \rightarrow \infty}\left[x-x^{2} \ln \left(\frac{1+x}{x}\right)\right]\\
&=\lim _{x \rightarrow \infty}\left[x-x^{2} \ln \left(\frac{1}{x}+1\right)\right]\\
& \,\ [ \text{Let} \,\ t= \frac{1}{x}, \text{so, as } x \rightarrow \infty, t \rightarrow \ 0^{+} . ]\\
&=\lim _{t \rightarrow 0^{+}}\left[\frac{1}{t}-\frac{1}{t^{2}} \ln (t+1)\right]\\
&=\lim _{t \rightarrow 0^{+}} \frac{t-\ln (t+1)}{t^{2}}\\
& \,\,\,\,\,[ \text{form is} \,\, 0 / 0\,\, \text {and by using L'Hôpital's rule we have} ]\\
&= \lim _{t \rightarrow 0^{+}} \frac{1-\frac{1}{t+1}}{2 t}\\
&=\lim _{t \rightarrow 0^{+}} \frac{t /(t+1)}{2 t}\\
&=\lim _{t \rightarrow 0^{+}} \frac{1}{2(t+1)}\\
&=\frac{1}{2}
\end{aligned}
$$
