New Notes Blog: What is the limit of (1+1/x)^(x+1/x) as x -> infinity ?

Wednesday, 19 March 2014

What is the limit of (1+1/x)^(x+1/x) as x -> infinity ?

This is an interesting limit. Lets define a variable y as:

$\displaystyle{y}=\lim_{{{x}\to\infty}}{{\left({1}+\frac{{1}}{{x}}\right)}}^{{{x}+\frac{{1}}{{x}}}}$

Lets take the logarithm of both sides.

$\displaystyle{\ln{{\left({y}\right)}}}=\lim_{{{x}\to\infty}}{\ln{{\left[{{\left({1}+\frac{{1}}{{x}}\right)}}^{{{x}+\frac{{1}}{{x}}}}\right]}}}$

$\displaystyle{\ln{{\left({y}\right)}}}=\lim_{{{x}\to\infty}}{\left({x}+\frac{{1}}{{x}}\right)}{\ln{{\left({1}+\frac{{1}}{{x}}\right)}}}$

Now I am going to make a quick substitution in the logarithm on the right hand side.

$\displaystyle{u}=\frac{{1}}{{x}}$

$\displaystyle{\ln{{\left({y}\right)}}}=\lim_{{{x}\to\infty}}{\left({x}+\frac{{1}}{{x}}\right)}{\ln{{\left({1}+{u}\right)}}}$

Now I will use a common taylor expansion, $\displaystyle{\ln{{\left({1}+{u}\right)}}}={\sum_{{{n}={0}}}^{\infty}}{{\left(-{1}\right)}}^{{{n}-{1}}}\cdot\frac{{{{u}}^{{n}}}}{{n}}={u}-\frac{{1}}{{2}}{{u}}^{{2}}+\frac{{1}}{{3}}{{u}}^{{3}}-\ldots$

$\displaystyle{\ln{{\left({y}\right)}}}=\lim_{{{x}\to\infty}}{\left({x}+\frac{{1}}{{x}}\right)}{\ln{{\left({1}+{u}\right)}}}$

$\displaystyle{\ln{{\left({y}\right)}}}=\lim_{{{x}\to\infty}}{\left({x}+\frac{{1}}{{x}}\right)}{\left({u}-\frac{{1}}{{2}}{{u}}^{{2}}+\frac{{1}}{{3}}{{u}}^{{3}}-\ldots\right)}$

Back substitute in to get in terms of x.

$\displaystyle{\ln{{\left({y}\right)}}}=\lim_{{{x}\to\infty}}{\left({x}+\frac{{1}}{{x}}\right)}{\left(\frac{{1}}{{x}}-\frac{{1}}{{{2}{{x}}^{{2}}}}+\frac{{1}}{{{3}{{x}}^{{3}}}}-\ldots\right)}$

$\displaystyle{\ln{{\left({y}\right)}}}=\lim_{{{x}\to\infty}}\ldots$

This is an interesting limit. Lets define a variable y as:

$\displaystyle{y}=\lim_{{{x}\to\infty}}{{\left({1}+\frac{{1}}{{x}}\right)}}^{{{x}+\frac{{1}}{{x}}}}$

Lets take the logarithm of both sides.

$\displaystyle{\ln{{\left({y}\right)}}}=\lim_{{{x}\to\infty}}{\ln{{\left[{{\left({1}+\frac{{1}}{{x}}\right)}}^{{{x}+\frac{{1}}{{x}}}}\right]}}}$

$\displaystyle{\ln{{\left({y}\right)}}}=\lim_{{{x}\to\infty}}{\left({x}+\frac{{1}}{{x}}\right)}{\ln{{\left({1}+\frac{{1}}{{x}}\right)}}}$

Now I am going to make a quick substitution in the logarithm on the right hand side.

$\displaystyle{u}=\frac{{1}}{{x}}$

$\displaystyle{\ln{{\left({y}\right)}}}=\lim_{{{x}\to\infty}}{\left({x}+\frac{{1}}{{x}}\right)}{\ln{{\left({1}+{u}\right)}}}$

$\displaystyle{\ln{{\left({y}\right)}}}=\lim_{{{x}\to\infty}}{\left({x}+\frac{{1}}{{x}}\right)}{\left({u}-\frac{{1}}{{2}}{{u}}^{{2}}+\frac{{1}}{{3}}{{u}}^{{3}}-\ldots\right)}$

Back substitute in to get in terms of x.

$\displaystyle{\ln{{\left({y}\right)}}}=\lim_{{{x}\to\infty}}{\left({x}+\frac{{1}}{{x}}\right)}{\left(\frac{{1}}{{x}}-\frac{{1}}{{{2}{{x}}^{{2}}}}+\frac{{1}}{{{3}{{x}}^{{3}}}}-\ldots\right)}$

$\displaystyle{\ln{{\left({y}\right)}}}=\lim_{{{x}\to\infty}}{\left[{x}{\left(\frac{{1}}{{x}}-\frac{{1}}{{{2}{{x}}^{{2}}}}+\frac{{1}}{{{3}{{x}}^{{3}}}}-\ldots\right)}+\frac{{1}}{{x}}{\left(\frac{{1}}{{x}}-\frac{{1}}{{{2}{{x}}^{{2}}}}+\frac{{1}}{{{3}{{x}}^{{3}}}}-\ldots\right)}\right]}$

$\displaystyle{\ln{{\left({y}\right)}}}=\lim_{{{x}\to\infty}}{\left[{\left({1}-\frac{{1}}{{{2}{x}}}+\frac{{1}}{{{3}{{x}}^{{2}}}}-\ldots\right)}+{\left(\frac{{1}}{{{x}}^{{2}}}-\frac{{1}}{{{2}{{x}}^{{3}}}}+\frac{{1}}{{{3}{{x}}^{{4}}}}-\ldots\right)}\right]}$