Question

What is `lim_(xto0^+) ((1/x)-((1)/(e^(x)-1)))`?

Answer 1

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

Explanation:

Sum the two terms:

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

The limit is now in the indeterminate form `0/0` so we can now apply l'Hospital's rule:

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

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

and as this is till in the form `0/0` a second time:

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

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

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

graph{1/x-1/(e^x-1) [-10, 10, -5, 5]}