Limit of function with natural logarithm I need help solving this

tripiverded9 2022-01-02 Answered
Limit of function with natural logarithm
I need help solving this problem. I tried L'hospital and rearranging but nothing worked.
\(\displaystyle\lim_{{{x}\rightarrow\infty}}{\left({\ln{{\left({1}+{\frac{{{1}}}{{{x}}}}\right)}}}-{\frac{{{1}}}{{{x}+{1}}}}\right)}\)

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Expert Answer

soanooooo40
Answered 2022-01-03 Author has 714 answers
Using L'Hospital rule we get
\(\displaystyle\lim_{{{x}\rightarrow\infty}}{\frac{{{\ln{{\left({1}+{\frac{{{1}}}{{{x}}}}\right)}}}-{\frac{{{1}}}{{{x}+{1}}}}}}{{{\frac{{{1}}}{{{x}^{{2}}}}}}}}=\lim_{{{x}\rightarrow\infty}}{\frac{{{\frac{{{x}}}{{{x}+{1}}}}\cdot{\left(-{\frac{{{1}}}{{{x}^{{2}}}}}\right)}+{\frac{{{1}}}{{{\left({x}+{1}\right)}^{{2}}}}}}}{{{\frac{{-{2}}}{{{x}^{{3}}}}}}}}=\)
\(\displaystyle\lim_{{{x}\rightarrow\infty}}{\frac{{-{x}-{1}+{x}}}{{{x}{\left({x}+{1}\right)}^{{2}}}}}\cdot{\frac{{{x}^{{3}}}}{{{2}}}}=\lim_{{{x}\rightarrow\infty}}{\frac{{{1}}}{{{\left({x}+{1}\right)}^{{2}}}}}\cdot{\frac{{{x}^{{2}}}}{{{2}}}}={\frac{{{1}}}{{{2}}}}\)
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Tiefdruckot
Answered 2022-01-04 Author has 2049 answers
write your limit in the form
\(\displaystyle\lim_{{{x}\rightarrow\infty}}{\frac{{{\ln{{\left({1}+{\frac{{{1}}}{{{x}}}}\right)}}}-{\frac{{{1}}}{{{x}+{1}}}}}}{{{\frac{{{1}}}{{{x}^{{2}}}}}}}}\)
and use L'Hospital
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Vasquez
Answered 2022-01-09 Author has 8850 answers

We have
\(x(\frac{\ln(1+\frac{1}{x})}{\frac{1}{x}})-\frac{x^2}{1+x}\)
Let \(x=\frac{1}{y}\). We have,
\(=\frac{1}{y}\frac{\ln(1+y)}{y}-\frac{1}{y^2+y}\)
This is,
\(\begin{array}{}=\frac{1}{y}\frac{\ln(1+y)}{y}-\frac{1}{y}+\frac{1}{y+1} \\=\frac{1}{y}(\frac{\ln(1+y)}{y}-1)+\frac{1}{y+1} \\=\frac{\ln(1+y)-y}{y^2}+\frac{1}{y+1} \end{array}\)
The first limit, as \(y\rightarrow0^+\) can be computed by LH used twice.
\(\rightarrow-\frac{1}{2}+1=\frac{1}{2}\)

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