How do I determine $\lim_{x\to\infty} \left[x - x^{2} \log\left(1 + 1/x\right)\right]$?

$$\begin{eqnarray*}x-x^2\log\frac{x+1}{x} &=& x-\int_{x}^{x+1}\frac{x^2}{t}\,dt\\&=&x-\int_{0}^{1}\frac{x^2}{t+x}\,dt\\&=&\int_{0}^{1}\frac{tx}{t+x}\,dt\\&=&\int_{0}^{1}\frac{t}{1+\frac{t}{x}}\,dt\end{eqnarray*} $$ hence by the dominated convergence theorem $$ \lim_{x\to +\infty}\left(x-x^2\log\frac{x+1}{x}\right)=\int_{0}^{1}t\,dt = \color{red}{\frac{1}{2}}.$$

Using Taylor expansions

Set $y=1/x$ and we're now looking at

$$\lim_{y\to 0}{1\over y}\left(1-{\log(1+y)\over y}\right)$$

Now in the neighbourhood of $0$, one has

$$\log(1+y)=y-{y^2\over 2}+o(y^2)$$

And so

$$1-{\log(1+y)\over y}={y\over 2}+o(y)$$

And the limit we're looking for is ${1\over 2}$

Using L'Hospital rule

The limit rewrites

$$\lim_{y\to 0}{1\over y^2}\left(y-\log(1+y)\right)$$

Indeterminate of the form $0/0$. By L'Hospital rule it is equal to

$$\lim_{y\to 0}{1\over 2y}\left(1-{1\over 1+y}\right)$$

Yet another indeterminate of the form $0/0$. L'Hospital again the limit is

$$\lim_{y\to 0}{1\over 2}{1\over \left(1+y\right)^2}={1\over 2}$$