I have tried to evaluate this limit:
$$\lim_{x\to \infty } \left(x(x+1) \log \left(\dfrac{x+1}{x} \right)-x\right)=\frac12$$
using $\lim_ {x\to \infty }\left(1+\dfrac{1}{x}\right)^{x}=e$, and using the variable change $z=\dfrac{1}{x}$ to get some known and standrad limit related to $\log$ natural logarithm properties function but I didn't succeed? Then any way and it's good if there is a suitable way for high school level.
Further to my comment, L'Hôpital's rule applied twice $$\lim\limits_{x\to\infty}\frac{(x+1)\log\left(\frac{x+1}{x}\right)-1}{\frac{1}{x}}= \lim\limits_{x\to\infty}\frac{\log\left(1 + \frac{1}{x}\right)-\frac{1}{x}}{-\frac{1}{x^2}}=\\ \lim\limits_{x\to\infty}\frac{\frac{1}{x^2 + x^3}}{\frac{2}{x^3}}= \lim\limits_{x\to\infty}\frac{1}{2}\cdot\frac{x^3}{x^2+x^3}=\frac{1}{2}$$ It is worth mentioning that both times we are dealing with $\frac{0}{0}$, so L'Hôpital's rule can be applied. L'Hôpital's rule used to be part of the high school program, I hope it still is.
Set $1/x=h\implies h\to0$
$$\lim_h\dfrac{(1+h)\ln(1+h)-h}{h^2}=\lim_h\dfrac{\ln(1+h)-h}h+\lim_h\dfrac{\ln(1+h)}h$$
For the first limit, use Are all limits solvable without L'Hôpital Rule or Series Expansion
What about the second one?
The second part is tricky to find without L'Hospital or Taylor. Since it's of the type $ \infty \cdot 0$. Obviously, it's equal to $-1/2$
– Yuriy S Aug 17 '19 at 19:40