$\lim\limits_{x \to \infty} {x}(a^{1/x}-1)$, such that $a > 1$

Question

$\lim\limits_{x \to \infty} {x}(a^{1/x}-1)$, such that $a > 1$

My Approach:

Since $x$ approaches infinity I presume we have to 'get rid' of the $x$ on the outside because the root $-1$ will be more than 0 but I have no idea how to reduce the $x$ so please help me out!

Answer 1

$$L=\lim_{x \to \infty} x(a^{1/x}-1)$$ let $x=1/t$, then $$L=\lim_{t\to 0} \frac{a^t-1}{t}=\lim_{t \to 0} \frac{1+t\ln a+(t\ln a)^2/2+...-1}{t} =\ln a.$$

Answer 2

Recall : $\lim_{y \rightarrow 0} \dfrac{e^y-1}{y}=1;$

Set $y:=1/x,$ and consider $y \rightarrow 0^+$.

$\lim_{y \rightarrow 0^+}\log a (\dfrac{e^{y\log a-1}}{y\log a}) = \log a. $

Answer 3

Here's an answer to a slightly more general limit

$ \displaystyle L(m, n) = \lim_{x \to \infty} x^{m} ( a^{\frac{1}{x^{n}}} -1) \tag*{} $

Where $ a>1 $ and $(m, n) $ are natural numbers.

$ \displaystyle L(m, n) = \lim_{x \to \infty} x^{m} ( e^{ \frac{\ln(a)}{x^n}} -1) = \lim_{x \to \infty} \sum_{r=1}^{\infty} \frac{x^{m-rn}}{r!}(\ln(a))^{r} \tag*{} $

Where we have used the series expansion of the exponential function in the step above. It should be now straight forward to observe that the given limit is equal to $0$ whenever $ m<n$, Equal to $\ln(a) $ when $m=n$ and the limit diverges to infinity whenever $m>n$.