Compute the limit: $\lim_{x\to\infty}\left(x-x^2\ln\frac{x+1}{x}\right)$

Question

$$\lim_{x\to\infty}\left(x-x^2\ln\frac{x+1}{x}\right)$$ What I did: $l=\lim_{y\to 0}\frac{y-\ln(y+1)}{y^2}$, putting $y=\frac{1}{x}$. The answer seems to be $1/2$ but how do I get to that?

Answer 1

By de l'Hôpital $$\lim_{y\to 0}\frac{y-\ln(y+1)}{y^2} =\lim_{y\to 0}\frac{1-\frac{1}{y+1}}{2y}=\lim_{y\to 0}\frac{1}{2(y+1)}.$$

Answer 2

Taylor:

$$\frac{y-\ln(y+1)}{y^2} = \frac{y - (y-\frac{y^2}{2} + O(y^3))}{y^2}=\frac{1}{2} +O(y) \stackrel{y\rightarrow 0^+}{\longrightarrow} \frac{1}{2}$$

Answer 3

By Taylor’s expansion

• $\ln\frac{x+1}{x}=\ln (1+1/x)=\frac 1x-\frac12x^2+o(x^{-2})$

then

$$x-x^2\ln\frac{x+1}{x}=\frac12+o(1)\to -\frac12$$

Answer 4

Use Taylor's formula at order $2$ for the log:$$\ln\frac{x+1}x=\ln\Bigl(1+\frac1x\Bigr)=\frac1x-\frac1{2x^2}+o\Bigl(\frac1{x^2}\Bigr),$$ so $$x-x^2\ln\frac{x+1}x=x-x^2\Bigl(\frac1x-\frac1{2x^2}+o\Bigl(\frac1{x^2}\Bigr)\Bigr)=x-x+\frac12-x^2o\Bigl(\frac1{x^2}\Bigr)=\frac12+o(1)\to\frac12.$$