# How can I prove this :$\lim_{x\to \infty } \left(x(x+1) \log \left(\dfrac{x+1}{x} \right)-x\right)=\frac12$ for high school level?

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.

• $$\log (\frac{x+1}{x} )=\log (1+\frac{1}{x} )= \frac{1}{x} -\frac{1}{2x^2} +\frac{1}{3x^3} - \ldots$$ when $x>1$ – Yuriy S Aug 17 at 19:09
• Thanks , but is there anyway available for high school student – zeraoulia rafik Aug 17 at 19:11
• But if you are not allowed to use Taylor series (an unfortunate, but real occurence) then I can try to work it out in another way – Yuriy S Aug 17 at 19:15
• Thanks youris , I want a sutiable way for high school level if it is possible , i have tried manytimes but no result – zeraoulia rafik Aug 17 at 19:16
• Can you use L'Hospital's Rule? – DonAntonio Aug 17 at 19:21

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 $$1/x=h\implies h\to0$$
$$\lim_h\dfrac{(1+h)\ln(1+h)-h}{h^2}=\lim_h\dfrac{\ln(1+h)-h}h+\lim_h\dfrac{\ln(1+h)}h$$