If $f(x) = {{\sqrt[4]{x^2\tan^2 4x}}\over{2x}}$
How can I evaluate $$\lim_{x\to 0^-} f(x)$$
Here's what I tried:
\begin{align} \lim_{x\to 0^-} {{\sqrt[4]{x^2\tan^2 4x}}\over{2x}} & = {1\over 2}\lim_{x\to 0^-} {{\sqrt[4]{x^2\tan^2 4x}}\over{x}} \\ & = {1\over 2}\lim_{x\to 0^-} {{\sqrt[4]{x^2\tan^2 4x}}\over{-\sqrt[4]{x^4}}} \\ & = -{1\over 2}\lim_{x\to 0^-} \sqrt[4]{{x^2\tan^2 4x}\over{x^4}} \end{align}
In the second step I wrote $x$ as $-\sqrt[4]{x^4}$ because we are working with the left limit which is on the left of $0$
\begin{align} -{1\over 2}\lim_{x\to 0^-} \sqrt[4]{{x^2\tan^2 4x}\over{x^4}} & = -{1\over 2}\lim_{x\to 0^-} \sqrt[4]{{\tan^2 4x}\over{x^2}} \\ & = -{1\over 2}\lim_{x\to 0^-} \sqrt[4]{\left({\tan 4x}\over{x}\right)^2} \\ & = -{1\over 2} \sqrt[4]{\left( \lim_{x\to 0^-} {{\tan 4x}\over{x}}\right)^2} \\ & = -{1\over 2} \sqrt[4]{(4)^2} \\ & = -{1\over 2} \sqrt[4]{16} \\ & = -{1\over 2} × 2 = -1 \end{align}
Are all these steps correct? I feel something wrong about the second step, because I found some other people who solved it $1$ and not $-1$ as my solution brought me.
I would write $$\lim_{x\to 0^-}\frac{\sqrt{|x\tan(4x)|}}{2x}$$