Question

What is the limit of x - x^2 * ln((1 + x) / x) when x approaches infinity?

Answer 1

\[\lim_{x\to\infty}\left[x - x^2 \cdot \ln\left(\frac{1 + x}{x}\right)\right]\]


Let \(\displaystyle u=\ln\left(\frac{1 + x}{x}\right)=\ln\left(\frac{1}{x}+1\right)\). Then \(\displaystyle x=\frac{1}{e^u-1}\).

Our limit becomes

\[\lim_{u\to 0^+}\left[\frac{1}{e^u-1}-\frac{1}{(e^u-1)^2}u\right]\]

\[=\lim_{u\to 0^+}\frac{e^u-1-u}{(e^u-1)^2}\]

Using L'Hospital's rule yields

\[=\lim_{u\to 0^+}\frac{e^u-1}{2(e^u-1)e^u}=\lim_{u\to 0^+}\frac{e^u-1}{2e^{2u}-2e^u}\]

Using L'Hospital's rule again

\[=\lim_{u\to 0^+}\frac{e^u}{4e^{2u}-2e^u}=\frac{1}{4-2}=\frac{1}{2}\]

Answer 2

Thanks :)