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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} $

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HINT:

Rationalize the D & N using $a^n-b^n=(a-b)(a^{n-1}+a^{n-2}b+\cdots+ab^{n-2}+b^{n-1})$

Then use $\lim_{u\to0}\dfrac{\arctan u}u=\lim_{y\to0}\dfrac y{\tan y}=1$ setting $\arctan u =y$

Similarly for $\arcsin$

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  • $\begingroup$ @Tim, $$a^{1/3}=u, a=u^3, b^{1/3}=v$$ etc. $\endgroup$ – lab bhattacharjee Feb 9 '16 at 6:14
  • $\begingroup$ I solved the problem, but since I've chosen your answer, could you go a little further and elaborate on how $(\arctan{x} + \arcsin{x})$ is asymptotically equivalent to $2x$ as $x\rightarrow0$? (not for me, but for future readers) $\endgroup$ – afsmi Feb 9 '16 at 8:55
  • $\begingroup$ @afsmi, $\lim_{u\to0}\dfrac{\arctan u}u=\lim_{y\to0}\dfrac y{\tan y}=1$ $\endgroup$ – lab bhattacharjee Feb 9 '16 at 8:59
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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 $\lim\limits_{x \to a}\dfrac{x^{n} - a^{n}}{x - a} = na^{n - 1}$.

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