If $\mathrm{L}=\underset{\mathrm{x}\to \mathrm{\infty }}{lim} \left\{\mathrm{x}-{\mathrm{x}}^2{\log }_{\mathrm{e}}\left(1+\frac{1}{\mathrm{x}}\right)\right\},$ then the value of 40L is ______ .

Let x = 1/y, Then
$\begin{array}{l}\underset{\mathrm{x}\to \mathrm{\infty }}{lim} \left(\mathrm{x}-{\mathrm{x}}^2{\log }_{\mathrm{e}}\left(1+\frac{1}{\mathrm{x}}\right)\right)=\underset{\mathrm{y}\to 0}{lim} \left(\frac{1}{\mathrm{y}}-\frac{{\log }_{\mathrm{e}}(1+\mathrm{y})}{{\mathrm{y}}^2}\right)\\ =\underset{\mathrm{y}\to 0}{lim} \left(\frac{\mathrm{y}-{\log }_{\mathrm{e}}(1+\mathrm{y})}{{\mathrm{y}}^2}\right)\\ =\underset{\mathrm{y}\to 0}{lim} \left(\frac{\mathrm{y}-\left(\mathrm{y}-\frac{{\mathrm{y}}^2}{2}\right)}{{\mathrm{y}}^2}\right)=1/2\end{array}$