Question

How do you find the limit of `(e^x-1)^(1/lnx)` as x approaches 0?

Answer 1

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

Explanation:

To avoid any complications in domain with regard to `ln(x)`, we will take the limit as `x->0^+`

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

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

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

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

(Because `e^x` is a continuous function, we have `lime^f(x) = e^(limf(x))`)

Then, it remains to evaluate `lim_(x->0^+)ln(e^x-1)/ln(x)`

As `lim_(x->0^+)ln(x)=-oo`, direct substitution leads to an `oo/oo` indeterminate form. Thus, we can apply L'Hopital's rule.

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

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

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

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

(As this leads to a `0/0` indeterminate form, we apply L'Hopital's rule once more.)

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

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

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

`=1`

Substituting back into the original problem, we find our answer:

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

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

`=e^1`

`=e`