Question

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

Answer 1

`Lim_(xrarr0^+)(1/x-1/(e^x-1))=1/2`

Explanation:

`Lim_(xrarr0^+)(1/x-1/(e^x-1))`

`Lim_(xrarr0^+)((e^x-1)-x)/(x(e^x-1))=0/0`

Using L'Hopitals rule:

`Lim_(xrarr0^+)(e^x-1)/((e^x-1)+xe^x)=0/0`

Again

`Lim_(xrarr0^+)(e^x)/(e^x+xe^x+e^x)=Lim_(xrarr0^+)cancel(e^x)/(cancel(e^x)(1+x+1)`

`Lim_(xrarr0^+)(1)/(x+2)=1/2`

Answer 2

`1/2`

Explanation:

`1/x - 1/(e^x - 1) = (e^x - x-1)/((e^x-1) x)`

This is a limit of type l'Hopital `0/0` so calling

`u(x) = e^x - x-1`
`v(x)=(e^x-1) x`

we have

`(d^2u)/(dx^2) =e^x`
`(d^2v)/(dx^2) =(2 +x)e^x` and then

`lim_(x->0)1/x - 1/(e^x - 1) = lim_(x->0)((d^2u)/(dx^2))/((d^2v)/(dx^2))=lim_(x->0)e^x/((2+x)e^x) = 1/2`