How do I determine $\lim_{x\to\infty} \left[x - x^{2} \log\left(1 + 1/x\right)\right]$? How do I determine this limit ?:
$$
\lim_{x \to \infty}\left[x - x^{2}\log\left(1 + {1 \over x}\right)\right]
$$
I have tried to decompose the $\log$ function but I don't know how to proceed from there. 
 A: $$\begin{eqnarray*}x-x^2\log\frac{x+1}{x} &=& x-\int_{x}^{x+1}\frac{x^2}{t}\,dt\\&=&x-\int_{0}^{1}\frac{x^2}{t+x}\,dt\\&=&\int_{0}^{1}\frac{tx}{t+x}\,dt\\&=&\int_{0}^{1}\frac{t}{1+\frac{t}{x}}\,dt\end{eqnarray*} $$
hence by the dominated convergence theorem
$$ \lim_{x\to +\infty}\left(x-x^2\log\frac{x+1}{x}\right)=\int_{0}^{1}t\,dt = \color{red}{\frac{1}{2}}.$$
A: 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
$$\lim_{y\to 0}{1\over 2}{1\over \left(1+y\right)^2}={1\over 2}$$
A: Hint:
Organise the limit as $\lim_{x \to \infty} (x - \frac{\log(1-1/x)}{1/x^2})$.
Can you see the answer now ?
A: The function is $\log(e^x)-\log(1+1/x)^{x^2}$ or $\log ((e^x) / (1+1/x)^{x^2})$. The logarithm goes to infinity as its argument goes to infinity so we only need to show its argument goes to infinity when x goes to infinity. But this is the quotient of a function that clearly goes to infinity with one that goes to 1 so it goes to infinity (as $x$ goes to infinity).
