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:

`y=lim_(x->oo) (1+1/x)^(x+1/x)`

Lets take the logarithm of both sides.

`ln(y)=lim_(x->oo) ln[(1+1/x)^(x+1/x)]`

`ln(y)=lim_(x->oo) (x+1/x) ln(1+1/x)`

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

`u=1/x`

`ln(y)=lim_(x->oo) (x+1/x) ln(1+u)`

Now I will use a common taylor expansion, `ln(1+u)=sum_(n=0)^oo (-1)^(n-1)*(u^n)/n=u-1/2u^2+1/3u^3-...`

`ln(y)=lim_(x->oo) (x+1/x) ln(1+u)`

`ln(y)=lim_(x->oo) (x+1/x) (u-1/2u^2+1/3u^3-...)`

Back substitute in to get in terms of x.

`ln(y)=lim_(x->oo) (x+1/x) (1/x-1/(2x^2)+1/(3x^3)-...)`

`ln(y)=lim_(x->oo)...

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

`y=lim_(x->oo) (1+1/x)^(x+1/x)`

Lets take the logarithm of both sides.

`ln(y)=lim_(x->oo) ln[(1+1/x)^(x+1/x)]`

`ln(y)=lim_(x->oo) (x+1/x) ln(1+1/x)`

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

`u=1/x`

`ln(y)=lim_(x->oo) (x+1/x) ln(1+u)`

Now I will use a common taylor expansion, `ln(1+u)=sum_(n=0)^oo (-1)^(n-1)*(u^n)/n=u-1/2u^2+1/3u^3-...`

`ln(y)=lim_(x->oo) (x+1/x) ln(1+u)`

`ln(y)=lim_(x->oo) (x+1/x) (u-1/2u^2+1/3u^3-...)`

Back substitute in to get in terms of x.

`ln(y)=lim_(x->oo) (x+1/x) (1/x-1/(2x^2)+1/(3x^3)-...)`

`ln(y)=lim_(x->oo) [x (1/x-1/(2x^2)+1/(3x^3)-...)+1/x(1/x-1/(2x^2)+1/(3x^3)-...)]`

`ln(y)=lim_(x->oo) [(1-1/(2x)+1/(3x^2)-...)+(1/x^2-1/(2x^3)+1/(3x^4)-...)]`

Now take the limit, every term that has an x in the denominator will go to zero leaving just one term.

`ln(y)=1`

Give both sides a new base e

`e^ln(y)=e^1`

`y=e`

Remember our original definition of y.

`y=lim_(x->oo) (1+1/x)^(x+1/x)=e`

This limit is equal to e.