# How do you find the limit of x(a^(1/x)-1) as x->oo?

Jun 1, 2017

${\log}_{e} a$

#### Explanation:

$x \left({a}^{\frac{1}{x}} - 1\right) = \frac{{a}^{0 + \frac{1}{x}} - {a}^{0}}{\frac{1}{x}}$ now calling $\frac{1}{x} = h$ we have

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

Jun 1, 2017

$\ln \left(a\right)$

#### Explanation:

Rewrite the limit as

${\lim}_{x \to \infty} \frac{{a}^{\frac{1}{x}} - 1}{\frac{1}{x}}$

So that it produces the indeterminate form $\frac{0}{0}$.

Now use L'hopital's rule (and chain rule as a result):

${\lim}_{x \to \infty} \frac{{a}^{\frac{1}{x}} - 1}{\frac{1}{x}} = {\lim}_{x \to \infty} \frac{{a}^{\frac{1}{x}} \cdot \ln \left(a\right) \cdot \frac{- 1}{x} ^ 2}{\frac{- 1}{x} ^ 2}$

$= {\lim}_{x \to \infty} {a}^{\frac{1}{x}} \ln \left(a\right) = {a}^{0} \ln \left(a\right) = \ln \left(a\right)$