# Limit of function with natural logarithm I need help solving this

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

• Questions are typically answered in as fast as 30 minutes

### Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

soanooooo40
Using L'Hospital rule we get
$$\displaystyle\lim_{{{x}\rightarrow\infty}}{\frac{{{\ln{{\left({1}+{\frac{{{1}}}{{{x}}}}\right)}}}-{\frac{{{1}}}{{{x}+{1}}}}}}{{{\frac{{{1}}}{{{x}^{{2}}}}}}}}=\lim_{{{x}\rightarrow\infty}}{\frac{{{\frac{{{x}}}{{{x}+{1}}}}\cdot{\left(-{\frac{{{1}}}{{{x}^{{2}}}}}\right)}+{\frac{{{1}}}{{{\left({x}+{1}\right)}^{{2}}}}}}}{{{\frac{{-{2}}}{{{x}^{{3}}}}}}}}=$$
$$\displaystyle\lim_{{{x}\rightarrow\infty}}{\frac{{-{x}-{1}+{x}}}{{{x}{\left({x}+{1}\right)}^{{2}}}}}\cdot{\frac{{{x}^{{3}}}}{{{2}}}}=\lim_{{{x}\rightarrow\infty}}{\frac{{{1}}}{{{\left({x}+{1}\right)}^{{2}}}}}\cdot{\frac{{{x}^{{2}}}}{{{2}}}}={\frac{{{1}}}{{{2}}}}$$
###### Not exactly what you’re looking for?
Tiefdruckot
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
Vasquez

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.
$$\rightarrow-\frac{1}{2}+1=\frac{1}{2}$$