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Question Number 65464 by aliesam last updated on 30/Jul/19

Commented by mathmax by abdo last updated on 30/Jul/19

$$letA\left(x\right)=x(x+1)ln\left(\frac{x+1}{x}\right)-xwehaveforx\in V(+\mathrm{\infty })$$$$ln\left(\frac{x+1}{x}\right)=ln(1+\frac{1}{x})=\frac{1}{x}-\frac{1}{2x^2}+o\left(\frac{1}{x^2}\right)$$$$A\left(x\right)=(x^2+x)(\frac{1}{x}-\frac{1}{2x^2}+o\left(\frac{1}{x^2}\right))-x$$$$=x-\frac{1}{2}+1-\frac{1}{2x}-x+o\left(\frac{1}{x}\right)=\frac{1}{2}-\frac{1}{2x}+o\left(\frac{1}{x}\right)\Rightarrow {lim}_{x\to +\mathrm{\infty }}A\left(x\right)=\frac{1}{2}$$

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