Using L'Hospital rule we get

\(\displaystyle\lim_{{{x}\rightarrow\infty}}{\frac{{{\ln{{\left({1}+{\frac{{{1}}}{{{x}}}}\right)}}}-{\frac{{{1}}}{{{x}+{1}}}}}}{{{\frac{{{1}}}{{{x}^{{2}}}}}}}}=\lim_{{{x}\rightarrow\infty}}{\frac{{{\frac{{{x}}}{{{x}+{1}}}}\cdot{\left(-{\frac{{{1}}}{{{x}^{{2}}}}}\right)}+{\frac{{{1}}}{{{\left({x}+{1}\right)}^{{2}}}}}}}{{{\frac{{-{2}}}{{{x}^{{3}}}}}}}}=\)

\(\displaystyle\lim_{{{x}\rightarrow\infty}}{\frac{{-{x}-{1}+{x}}}{{{x}{\left({x}+{1}\right)}^{{2}}}}}\cdot{\frac{{{x}^{{3}}}}{{{2}}}}=\lim_{{{x}\rightarrow\infty}}{\frac{{{1}}}{{{\left({x}+{1}\right)}^{{2}}}}}\cdot{\frac{{{x}^{{2}}}}{{{2}}}}={\frac{{{1}}}{{{2}}}}\)

\(\displaystyle\lim_{{{x}\rightarrow\infty}}{\frac{{{\ln{{\left({1}+{\frac{{{1}}}{{{x}}}}\right)}}}-{\frac{{{1}}}{{{x}+{1}}}}}}{{{\frac{{{1}}}{{{x}^{{2}}}}}}}}=\lim_{{{x}\rightarrow\infty}}{\frac{{{\frac{{{x}}}{{{x}+{1}}}}\cdot{\left(-{\frac{{{1}}}{{{x}^{{2}}}}}\right)}+{\frac{{{1}}}{{{\left({x}+{1}\right)}^{{2}}}}}}}{{{\frac{{-{2}}}{{{x}^{{3}}}}}}}}=\)

\(\displaystyle\lim_{{{x}\rightarrow\infty}}{\frac{{-{x}-{1}+{x}}}{{{x}{\left({x}+{1}\right)}^{{2}}}}}\cdot{\frac{{{x}^{{3}}}}{{{2}}}}=\lim_{{{x}\rightarrow\infty}}{\frac{{{1}}}{{{\left({x}+{1}\right)}^{{2}}}}}\cdot{\frac{{{x}^{{2}}}}{{{2}}}}={\frac{{{1}}}{{{2}}}}\)