Transforming the given expression into a form that can use the L'Hopital Rule

$$\lim_{x\to\infty}\left[x-x^2\ln\left(\frac{1+x}{x}\right)\right]=\lim_{n\to\infty}\frac{\frac{1}{x}-\ln\left(\frac{1+x}{x}\right)}{\frac{1}{x^2}}=\cdots=\frac{1}{2}$$