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

2 Answers
Dec 23, 2017

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

Dec 23, 2017

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