Show that $\lim \limits_{x \to \infty} (x + x^2 \log \frac{x}{x+1}) = \frac{1}{2}$ I am self studying real analysis and I found this problem in a book.
We have to show that if $x > 1$ then
$$\lim_{x \to \infty} \left[ x + x^2 \log \frac{x}{x+1} \right] = \frac{1}{2}$$.
I can prove that the sequence is bounded.
It is bounded below by 0.
Suppose not, then there exists a $x > 1$ such that
$$x + x^2 \log \frac{x}{x + 1} < 0$$.
Which in turn implies
$$1 + \frac{1}{x} > e^{\frac{1}{x}}$$,
a contradiction.
It is bounded above by 1.
Suppose not, then there exists a $x > 1$ such that
$$x + x^2 \log \frac{x}{x + 1} > 1$$.
Which means $1 + \frac{1}{x} < e^{\frac{1}{x} - \frac{1}{x^2}}$.
Which in turn implies
$$\frac{1}{x^2} < \sum_{n = 2}^{\infty} \frac{\left( \frac{1}{x} - \frac{1}{x^2} \right)^n}{n!}$$.
Now,
$$\sum_{n = 2}^{\infty} \frac{\left( \frac{1}{x} - \frac{1}{x^2} \right)^n}{n!} < \sum_{n = 2}^{\infty} \left( \frac{1}{x} - \frac{1}{x^2} \right)^n = \left( \frac{1}{x} - \frac{1}{x^2} \right)^2 \sum_{n = 0}^{\infty} \left( \frac{1}{x} - \frac{1}{x^2} \right)^n = \frac{1}{x^2} \frac{x^2 - 2x + 1}{x^2 - x + 1} < \frac{1}{x^2}$$a contradiction.
However the sequence is not monotonic.
Let $u(x) = x + x^2 \log \frac{x}{x+1}$.
Then $u(183073) \approx 0.5000000077125151$, $u(183074) \approx 0.4999997131235432$ and $u(183089) \approx 0.5000000357104$
Any help on this?
Thanks.

Set $1/x=h$
to get $$\lim_{h\to0}\dfrac{h-\ln(1+h)}{h^2}$$
Now use $\ln(1+h)=h-\dfrac{h^2}2+O(h^3)$
