Finding the limit $\lim_{x\to+\infty}(x-x^2\log(1+1/x))$ in a elementary way How to find the limit $$\lim_{x\to+\infty}\left(x-x^2\log\left(1+\frac{1}{x}\right)\right)$$  in a elementary way? 
  I can solve with Taylor expansion, but it is placed in the beginning of my calculus book, so I should only use things like:
-Main theorems involving limits, including the limits for $x\to 0$, $\lim\frac{\sin x}{x}$, $\lim\frac{e^x-1}{x}$, $\lim\frac{\log(x+1)}{x}$, $\lim\frac{(x+1)^p-1}{x}$
-$\frac{x}{1+x} \leq \log(1+x) \leq x$ or similar inequalities
I cannot use derivatives, Taylor expansion, $o(x), O(x)$ and similar things.
Using the inequality that I have written above and the substitution $x=\frac{1}{\sin t}$ I have been only able to prove that the limit is greater or equal than 0 and smaller or equal than 1.
Any ideas?
 A: Let $t = 1/x$ so that $t \to 0^{+}$ as $x \to \infty$. Then the desired limit is $\lim\limits_{t \to 0^{+}}\dfrac{t - \log(1 + t)}{t^{2}}$. If we have $y > 0$ then we know that $1 - y^{2} < 1 < 1 + y^{3}$ and dividing by $(1 + y)$ we get $$1 - y < \dfrac{1}{1 + y} < 1 - y + y^{2}$$ Integrating (with respect to $y$) in the interval $[0, t]$ we get $$t - \frac{t^{2}}{2} < \log(1 + t) < t - \frac{t^{2}}{2} + \frac{t^{3}}{3}$$ for $t > 0$. this means that $$ \frac{1}{2} - \frac{t}{3} < \frac{t - \log(1 + t)}{t^{2}} < \frac{1}{2}$$ for $t > 0$. Now using squeeze theorem we get the desired limit as $1/2$. Note that I have used integration to establish the logarithmic inequalities. The inequalities can be established by using any other definition of $\log x$ also, but the definition based on integrals seems to be the easiest to deal with.
Update: Based on OP comment I think it is better to transform the limit into one involving exponential functions. Let $1 + t = e^{z}$ so that $ t \to 0^{+}$ implies $z \to 0^{+}$. Then we have $$\begin{aligned}L &= \lim_{t \to 0^{+}}\frac{t - \log(1 + t)}{t^{2}}\\
&= \lim_{z \to 0^{+}}\frac{e^{z} - 1 - z}{(e^{z} - 1)^{2}}\\
&= \lim_{z \to 0^{+}}\frac{e^{z} - 1 - z}{z^{2}}\cdot\left(\frac{z}{e^{z} - 1}\right)^{2}\\
&= \lim_{z \to 0^{+}}\frac{e^{z} - 1 - z}{z^{2}}\end{aligned}$$ Now there are two ways. First the easier one as follows. We have $$\begin{aligned}\frac{e^{z} - 1 - z}{z^{2}} &= \frac{1}{2!} + \frac{z}{3!} + \frac{z^{2}}{4!} + \cdots\\
&= \frac{1}{2} + \phi(z)\end{aligned}$$ where $$\phi(z) = \frac{z}{3!} + \frac{z^{2}}{4!} + \cdots$$ so that $$0 < \phi(z) \leq \frac{z}{6} + \frac{z^{2}}{18} + \frac{z^{3}}{54} + \cdots$$ i.e $$0 < \phi(z) \leq \frac{z/6}{1 - (z/3)} = \frac{z}{6 - 2z}$$ for $0 < z < 3$. By Squeeze theorem as $z \to 0^{+}$ we have $\phi(z) \to 0$. Hence $(e^{z} - 1 - z)/z^{2} \to 1/2$.
The hard part is to show that $(e^{z} - 1 - z)/z^{2} \to 1/2$ without using the series for $e^{z}$ and instead using the definition $e^{z} = \lim_{n \to \infty}(1 + z/n)^{n}$ directly. Let me know if you are interested in the hard version.
A: Here is how. We have 
$$ L =  \lim_{x\to \infty} x^2\left( \frac{1}{x} - \ln\left(1+\frac{1}{x}\right) \right). $$
Letting $y=\frac{1}{x}$ results in the form
$$ L = \lim_{y\to 0} \frac{1}{y^2}(y-\ln(1+y)) = -\lim_{y\to 0 } \frac{\frac{\ln(1+y)}{y}-1}{y-0}= -\lim_{y\to 0 } \frac{f(y)-1}{y-0}, $$
where 

$$ f(y) = \frac{\ln(1+y)}{y}. $$

I leave it here for you.
A: Hint:
$$x\log \bigg(1+\frac{1}{x}\bigg)=\log \bigg(1+\frac{1}{x}\bigg)^x$$
