I'm asked to find the value of this limit:
$$\lim_{x\to\ 0^+} \arccos(\ln(1-2x)^{\frac{1}{4x}})$$
I am NOT allowed to use L'hopital's rule here.
My initial goal was try to first rewrite the limit as:
$$\arccos(\lim_{x\to\ 0^+}\ln(1-2x)^{\frac{1}{4x}})$$
The I was trying to use the fact that:
$$\lim_{x\to\ \infty }(1+\frac{1}{x})^x=\ e$$
So I made the substitution $x=\frac{-1}{2u}$ so therefore, $-2u=\frac{1}{x}$ so I get that as $x$ approaches $0$, $u$ approaches $\infty$. So overall, my limit becomes:
$$\arccos(\lim_{x\to\ 0^+}\ln(1+\frac{1}{u})^{-2u})^{\frac{1}{4}}$$
However, now, I become stuck. Is this method even the right approach? How would I even begin to go about solving this?
For reference, Here is the question:
You can certainly pull out the arccosine, provided that $$ \lim_{x\to\ 0^+}\ln(1-2x)^{1/(4x)}\tag{*} $$ exists and belongs to $[-1,1]$.
Now I assume the exponential applies to $1-2x$ and so you can pull out also the logarithm, provided $$ \lim_{x\to\ 0^+}(1-2x)^{1/(4x)} $$ exists and is positive. Do the substitution $2x=1/t$ and the limit becomes $$ \lim_{t\to\infty}\biggl(\biggl(1-\frac{1}{t}\biggr)^{\!t}\biggr)^{\!1/2} $$ Can you go on from here?
The limit (*) can't be interpreted as $$ \lim_{x\to\ 0^+}\bigl(\ln(1-2x)\bigr)^{1/(4x)} $$ because the basis would be negative.