<s>[INST] How can I evaluate the left limit of this function? 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.
 [/INST] I would write $$\lim_{x\to 0^-}\frac{\sqrt{|x\tan(4x)|}}{2x}$$
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