How do I prove this limit doesn't exist without L'Hopital's Rule?

Question

I'm trying to prove that $$ \lim_{x \to 0+}\arccos\left(\ln(1-2x)^\frac{1}{4x}\right) $$ does not exist. I have already found out that $f(0)$ is $\pi/2$, and that the function only exists in Quadrant 2. Since the function only exists in Quadrant 2, the limit approaching $0$ from the right does not exist. I'm having trouble writing this out in words, and proving it using the limit. Is it as simple as just saying that, as ${x \to 0+}$, $\ln(1-2x)$ approaches $\ln(1)$, and the exponent $\frac{1}{4x}$ approaches $\frac{1}{0}$, meaning that the function is undefined because it gives us $0^\frac{1}{0}$?

Answer 1

$$\lim_{x\to 0^+} \, \arccos(\log (1-2 x))^{1/4x}$$ Cannot exist for another reason

The domain of the function must guarantee the existence of logarithm and arccos

Therefore must be $$0<(\log (1-2 x))^{1/4x}\leq 1$$ taking logarithm of both sides $$\frac{1}{4x}\log(\log(1-2x))\leq 0$$ $$x\leq \frac{1-e}{2}\approx-0.86$$ so the domain of this function is $\left(-\infty,\;\dfrac{1-e}{2}\right]$

no limit (right or left) can be computed for $x\to 0$

Hope this helps