How do I evaluate this without using taylor expansion :$\lim_{x \to \infty}x^2\log(\frac {x+1}{x})-x\ $? 
How do I evaluate this without using Taylor expansion?
$$\lim_{x \to \infty}x^2\log\left(\frac {x+1}{x}\right)-x$$

Note: I used Taylor expansion at $z=0$ and I have got $\frac{-1}{2}$
Thank you for any help
 A: An approach is to set $u:=\dfrac1x$, then you are looking for
$$
\lim_{u \to 0}\left(\frac{\log (1+u)-u}{u^2}\right)
$$ then you may conclude with L'Hospital's rule
$$
\lim_{u \to 0}\left(\frac{\log (1+u)-u}{u^2}\right)=\lim_{u \to 0}\left(\frac{\frac1{1+u}-1}{2u}\right)=\lim_{u \to 0}\left(\frac{\frac{-1}{(1+u)^2}}{2}\right)=-\frac12.
$$
A: Another option is to use the following (very convenient) inequality:
$$
\frac{2x}{2+x} \leq \ln(1+x) \leq \frac{x}{\sqrt{1+x}}
$$
which holds for $x \geq 0$. In your case, you have $\ln(1+\frac{1}{x})$, leading to
$$
x^2\frac{\frac{2}{x}}{2+\frac{1}{x}} -x \leq x^2\ln\!\left(1+\frac{1}{x}\right) -x \leq x^2\frac{\frac{1}{x}}{\sqrt{1+\frac{1}{x}}} - x.
$$


*

*For the LHS:
$$
x^2\frac{\frac{2}{x}}{2+\frac{1}{x}} -x = x\left(\frac{2}{2+\frac{1}{x}} - 1\right) = x\left(\frac{-\frac{1}{x}}{2+\frac{1}{x}}\right) = \frac{-1}{2+\frac{1}{x}} \xrightarrow[x\to\infty]{} -\frac{1}{2}.
$$

*And the RHS:
$$\begin{align}
x^2\frac{\frac{1}{x}}{\sqrt{1+\frac{1}{x}}} - x
&= x\left(\frac{1}{\sqrt{1+\frac{1}{x}}} - 1\right)
= x\left(\frac{1-\sqrt{1+\frac{1}{x}}}{\sqrt{1+\frac{1}{x}}}\right)\\
&= x\left(\frac{-\frac{1}{x}}{\sqrt{1+\frac{1}{x}}\left(1+\sqrt{1+\frac{1}{x}}\right)}\right)\\
&= \frac{-1}{\sqrt{1+\frac{1}{x}}\left(1+\sqrt{1+\frac{1}{x}}\right)}
 \xrightarrow[x\to\infty]{} -\frac{1}{2}.
\end{align}$$
It only remains to invoke the squeeze theorem.
(I also strongly suggest bookmarking the "useful inequalities sheet" linked at the beginning.)
