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

Jun 16, 2016

${\lim}_{x \to {0}^{+}} {\left({e}^{x} - 1\right)}^{\frac{1}{\ln} \left(x\right)} = e$

#### Explanation:

To avoid any complications in domain with regard to $\ln \left(x\right)$, we will take the limit as $x \to {0}^{+}$

${\lim}_{x \to {0}^{+}} {\left({e}^{x} - 1\right)}^{\frac{1}{\ln} \left(x\right)}$

$= {\lim}_{x \to {0}^{+}} {e}^{\ln} \left({\left({e}^{x} - 1\right)}^{\frac{1}{\ln} \left(x\right)}\right)$

$= {\lim}_{x \to {0}^{+}} {e}^{\ln \frac{{e}^{x} - 1}{\ln} \left(x\right)}$

$= {e}^{{\lim}_{x \to {0}^{+}} \ln \frac{{e}^{x} - 1}{\ln} \left(x\right)}$

(Because ${e}^{x}$ is a continuous function, we have $\lim {e}^{f} \left(x\right) = {e}^{\lim f \left(x\right)}$)

Then, it remains to evaluate ${\lim}_{x \to {0}^{+}} \ln \frac{{e}^{x} - 1}{\ln} \left(x\right)$

As ${\lim}_{x \to {0}^{+}} \ln \left(x\right) = - \infty$, direct substitution leads to an $\frac{\infty}{\infty}$ indeterminate form. Thus, we can apply L'Hopital's rule.

${\lim}_{x \to {0}^{+}} \ln \frac{{e}^{x} - 1}{\ln} \left(x\right)$

$= {\lim}_{x \to {0}^{+}} \frac{\frac{d}{\mathrm{dx}} \ln \left({e}^{x} - 1\right)}{\frac{d}{\mathrm{dx}} \ln \left(x\right)}$

$= {\lim}_{x \to {0}^{+}} \frac{{e}^{x} / \left({e}^{x} - 1\right)}{\frac{1}{x}}$

$= {\lim}_{x \to {0}^{+}} \frac{x {e}^{x}}{{e}^{x} - 1}$

(As this leads to a $\frac{0}{0}$ indeterminate form, we apply L'Hopital's rule once more.)

$= {\lim}_{x \to {0}^{+}} \frac{\frac{d}{\mathrm{dx}} \left(x {e}^{x}\right)}{\frac{d}{\mathrm{dx}} \left({e}^{x} - 1\right)}$

$= {\lim}_{x \to {0}^{+}} \frac{x {e}^{x} + {e}^{x}}{e} ^ x$

$= {\lim}_{x \to {0}^{+}} \left(x + 1\right)$

$= 1$

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

${\lim}_{x \to {0}^{+}} {\left({e}^{x} - 1\right)}^{\frac{1}{\ln} \left(x\right)}$

$= {e}^{{\lim}_{x \to {0}^{+}} \ln \frac{{e}^{x} - 1}{\ln} \left(x\right)}$

$= {e}^{1}$

$= e$