Find $\lim_{x\to 0}\left(1-a^x\right)^\frac 1{x}\quad (0<a<1)$

Question

Somebody asked $$\lim_{x\to\infty}\left(1-a^x\right)^\frac 1{x} \quad (0<a<1)$$ which is $1$. But the original question was missing the part $x\to\infty$, so I misinterpreted it as $x\to0^+$

$$\lim_{x\to 0^+}\left(1-a^x\right)^\frac 1{x} \quad (0<a<1)$$

which I couldn't yet figure out how to solve it. Wolfram says it's zero. Can somebody help? (without L'hopital if possible)

Answer 1

Write

$$\left(1-a^x\right)^{1/x}=e^{\frac1x\log(1-a^x)}$$

and now:

$$\lim_{x\to 0^+}\frac1x\log(1-a^x)=-\infty\implies \lim_{x\to 0^+}\left(1-a^x\right)^{1/x}=0$$

Observe that when $\;x<0,\,0<a<1\;$ , then $\;a^x>1\;$ so $\;\log(1-a^x)\;$ isn't defined as a real function

Answer 2

Hint: Take $\log$ and use L'Hôpital's rule.

Answer 3

When $x\rightarrow 0^+$ then $a^x\rightarrow 1^-$, thus $1-a^x\rightarrow 0^+$ and (when $0<x<1$) then $(1-a^x)^{1/x}\leq 1-a^x\rightarrow 0$. No need to use special tools.

[No longer needed: When $x\rightarrow 0^-$ there is no limit. We have $1-a^x\rightarrow 0^-$. If we want to stay real suppose $1/x=-n$ a negative integer. But then $(1-a^{1/n})^{-n} \rightarrow \pm \infty$ depending on $n$ being diverging along even or odd integers.]