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

Question

15444 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-06-16T08:02:15

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

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:

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

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

$=e^1$

$=e$

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