Question

`lim_(x->0)( (1+x)^(1/x) - e ) / (x) =` ?

Answer 1

`-e/2`

Explanation:

Developing `f(x)= (1+x)^(1/x)` in Taylor series for `x->0` we have

`f(x)=f(0)+f'(0)x+1/(2!)f''(0)x^2 + cdots`

Here

`f(0) = lim_(x->0)(1+x)^(1/x) = e`
`f'(0) = lim_(x->0)(1 + x)^(1/x) (1/(x (1 + x)) - Log(1 + x)/x^2)=`
`=lim_(x->0)(1+x)^(1/x)lim_(x->0)(1/(x (1 + x)) - Log(1 + x)/x^2)`

but applying l'Hopital's rules

`lim_(x->0)(1/(x (1 + x)) - Log(1 + x)/x^2)=lim_(x->0)(-1/(x+1))/(6x+2)=-1/2`

so

`f'(0) = -e/2`

At this point we conclude that the Taylor series develop as

`f(x) = e -e/2x+eC_2x^2+ cdots` with `C_2` a suitable constant.

Following we have now

`lim_(x->0)( (1+x)^(1/x) - e ) / (x) equiv lim_(x->0)(f(x)-e)/x =`

`lim_(x->0)(e -e/2x+eC_2x^2+ cdots-e)/x = -e/2`

Answer 2

`-e/2`

Explanation:

Knowing that `lim_(xrarr0)(1+x)^(1/x)=e`, we can see that this limit is in the indeterminate form `0/0`. This means we can use l'Hopitals rule to find the limit by taking the derivatives of the numerator and denominator separately.

`lim_(xrarr0)((1+x)^(1/x)-e)/x=lim_(xrarr0)(d/dx(1+x)^(1/x))/1`

Use logarithmic differentiation to find the derivative of `(1+x)^(1/x)`:

`y=(1+x)^(1/x)`

`ln(y)=1/xln(1+x)`

`1/y*dy/dx=-1/x^2ln(1+x)+1/(x(1+x))`

`dy/dx=(1+x)^(1/x)(1/(x(1+x))-ln(1+x)/x^2)`

So the limit is:

`=lim_(xrarr0)(1+x)^(1/x)(1/(x(1+x))-ln(1+x)/x^2)`

`=lim_(xrarr0)(1+x)^(1/x)((x-(1+x)ln(1+x))/(x^2(1+x)))`

We can move `lim_(xrarr0)(1+x)^(1/x)` out of the limit as `e`, but to work with the remaining portion of the limit we will use l'Hopital's again since we have `0/0`:

`=elim_(xrarr0)(d/dx(x-(1+x)ln(1+x)))/(d/dx(x^2+x^3))`

`=elim_(xrarr0)(1-ln(1+x)-(1+x)/(1+x))/(2x+3x^2)`

`=elim_(xrarr0)(-ln(1+x))/(2x+3x^2)`

Since we have `0/0`, use l'Hopital's again:

`=elim_(xrarr0)(-1/(1+x))/(2+6x)`

Which we finally can evaluate:

`=e((-1/1)/2)`

`=-e/2`