Evaluate this limit of inverse trigonometric and radical functions without l'Hospital How can I solve this using only 'simple' algebraic tricks and asymptotic equivalences? No l'Hospital.
$$\lim_{x \rightarrow0}
\frac
{\sqrt[3]{1+\arctan{3x}} - \sqrt[3]{1-\arcsin{3x}}}
{\sqrt{1-\arctan{2x}} - \sqrt{1+\arcsin{2x}}}
$$
Rationalizing the numerator and denominator gives
$$
\lim_{x \rightarrow0}
\frac
{A(\arctan{3x}+\arcsin{3x})}
{B(\arctan{2x} + \arcsin{2x})}
$$
where $\lim_{x \rightarrow 0} \frac{A}{B} = -\frac{2}{3} $

We will use the following standard limits $$\lim_{x \to 0}\frac{\arctan x}{x} = 1 = \lim_{x \to 0}\frac{\arcsin x}{x},\,\lim_{x \to a}\frac{x^{n} - a^{n}}{x - a} = na^{n - 1}\tag{1}$$ We have
\begin{align}
L &= \lim_{x \to 0}\frac{\sqrt[3]{1 + \arctan 3x} - \sqrt[3]{1 - \arcsin 3x}}
{\sqrt{1 - \arctan 2x} - \sqrt{1 + \arcsin 2x}}\notag\\
&= \lim_{x \to 0}\dfrac{\dfrac{\sqrt[3]{1 + \arctan 3x}}{x} - \dfrac{\sqrt[3]{1 - \arcsin 3x}}{x}}
{\dfrac{\sqrt{1 - \arctan 2x}}{x} - \dfrac{\sqrt{1 + \arcsin 2x}}{x}}\notag\\
&= \lim_{x \to 0}\dfrac{\left(\dfrac{\sqrt[3]{1 + \arctan 3x} - 1}{x}\right) - \left(\dfrac{\sqrt[3]{1 - \arcsin 3x} - 1}{x}\right)}
{\left(\dfrac{\sqrt{1 - \arctan 2x} - 1}{x}\right) - \left(\dfrac{\sqrt{1 + \arcsin 2x} - 1}{x}\right)}\notag\\
&= \frac{A - B}{C - D}\notag
\end{align}
The limit of each of the 4 bracketed expressions can be evaluated easily. These limits I have denoted by $A, B, C, D$ respectively. I will illustrate the complete evaluation for the first expression. We have
\begin{align}
A &= \lim_{x \to 0}\frac{\sqrt[3]{1 + \arctan 3x} - 1}{x}\notag\\
&= \lim_{x \to 0}\frac{\sqrt[3]{1 + \arctan 3x} - 1}{\arctan 3x}\cdot\frac{\arctan 3x}{3x}\cdot 3\notag\\
&= 3 \lim_{x \to 0}\frac{\sqrt[3]{1 + \arctan 3x} - 1}{\arctan 3x}\notag\\
&= 3 \lim_{t \to 1}\frac{t^{1/3} - 1}{t - 1}\text{ (putting }t = 1 + \arctan 3x)\notag\\
&= 3\cdot\frac{1}{3}\notag\\
&= 1\notag
\end{align}
Similarly $B = -1, C = -1, D = 1$ and hence the desired limit $L = (A - B)/(C - D) = -1$.
Rationalization of expressions involving radicals makes sense mostly if we are dealing with square roots. In cases where we have cube roots (and higher roots) it is difficult to write/type the long expressions (obtained during rationalization process) and it is best to make use of the standard limit formula $\lim\limits_{x \to a}\dfrac{x^{n} - a^{n}}{x - a} = na^{n - 1}$.
