<s>[INST] Compute limit of a function Compute: $$\lim_{x \rightarrow 0^+} \frac{\arctan(e^x+\arctan x)-\arctan(e^{\sin x}+\arctan(\sin x))}{x^3}$$ 
WolframAlpha tells me it's 1/6. Any nice idea how to rewrite that expression? Thanks! 
 [/INST] We can proceed as follows $$\begin{aligned}L\, &= \lim_{x \to 0^{+}}\frac{\tan^{-1}(e^{x} + \tan^{-1}x) - \tan^{-1}(e^{\sin x} + \tan^{-1}(\sin x))}{x^{3}}\\
&= \lim_{x \to 0^{+}}\dfrac{\tan^{-1}\left(\dfrac{e^{x} + \tan^{-1}x - e^{\sin x} - \tan^{-1}(\sin x)}{1 + (e^{x} + \tan^{-1}x)(e^{\sin x} + \tan^{-1}(\sin x))}\right)}{x^{3}}\\
&= \lim_{t \to 0^{+}}\dfrac{\tan^{-1}t}{t}\cdot\dfrac{t}{x^{3}}\\
&= \lim_{x \to 0^{+}}\frac{t}{x^{3}}\\
&= \lim_{x \to 0^{+}}\dfrac{\left(\dfrac{e^{x} + \tan^{-1}x - e^{\sin x} - \tan^{-1}(\sin x)}{1 + (e^{x} + \tan^{-1}x)(e^{\sin x} + \tan^{-1}(\sin x))}\right)}{x^{3}}\\
&= \frac{1}{2}\lim_{x \to 0^{+}}\frac{e^{x} + \tan^{-1}x - e^{\sin x} - \tan^{-1}(\sin x)}{x^{3}}\\
&= \frac{1}{2}\left(\lim_{x \to 0^{+}}\frac{e^{x} - e^{\sin x}}{x^{3}} + \frac{\tan^{-1}x - \tan^{-1}(\sin x)}{x^{3}}\right)\\
&= \frac{1}{2}\left(\lim_{x \to 0^{+}}\frac{e^{\sin x}(e^{x - 
\sin x} - 1)}{x^{3}} + \dfrac{\tan^{-1}\left(\dfrac{x - \sin x}{1 + x\sin x}\right)}{x^{3}}\right)\\
&= \frac{1}{2}\left(\lim_{x \to 0^{+}}\frac{e^{\sin x}(e^{x - 
\sin x} - 1)}{x^{3}} + \lim_{x \to 0^{+}}\dfrac{\tan^{-1}\left(\dfrac{x - \sin x}{1 + x\sin x}\right)}{x^{3}}\right)\\
&= \frac{1}{2}\left(\lim_{x \to 0^{+}}\frac{(e^{x - 
\sin x} - 1)}{x^{3}} + \lim_{v \to 0^{+}}\dfrac{\tan^{-1}v}{v}\cdot\frac{v}{x^{3}}\right)\\
&= \frac{1}{2}\left(\lim_{u \to 0^{+}}\frac{(e^{u} - 1)}{u}\cdot\frac{u}{x^{3}} + \lim_{v \to 0^{+}}\dfrac{\tan^{-1}v}{v}\cdot\frac{v}{x^{3}}\right)\\
&= \frac{1}{2}\left(\lim_{x \to 0^{+}}\frac{x - \sin x}{x^{3}} + \lim_{x \to 0^{+}}\frac{x - \sin x}{(1 + x\sin x)x^{3}}\right)\\
&= \frac{1}{2}\left(\lim_{x \to 0^{+}}\frac{x - \sin x}{x^{3}} + \lim_{x \to 0^{+}}\frac{x - \sin x}{x^{3}}\right)\\
&= \lim_{x \to 0^{+}}\frac{x - \sin x}{x^{3}}\\
&= \lim_{x \to 0^{+}}\frac{1 - \cos x}{3x^{2}}\text{ (by L'Hospital's Rule)}\\
&= \lim_{x \to 0^{+}}\frac{1}{3}\cdot\frac{2\sin^{2}(x/2)}{(x/2)^{2}}\cdot\frac{(x/2)^{2}}{x^{2}}\\
&= \frac{1}{3}\cdot 2\cdot\frac{1}{4} = \frac{1}{6}\end{aligned}$$ We have used subsitutions $$t = \dfrac{e^{x} + \tan^{-1}x - e^{\sin x} - \tan^{-1}(\sin x)}{1 + (e^{x} + \tan^{-1}x)(e^{\sin x} + \tan^{-1}(\sin x))}, u = x - \sin x, v = \dfrac{x - \sin x}{1 + x\sin x}$$ and each of these variables $t, u, v$ tends to $0$ as $x \to 0^{+}$ and hence we have $$\lim_{t \to 0^{+}}\frac{\tan^{-1}t}{t} = 1,\,\lim_{u \to 0^{+}}\frac{e^{u} - 1}{u} = 1,\, \lim_{v \to 0^{+}}\frac{\tan^{-1}v}{v} = 1$$ The above derivation shows that the normal rules of algebra of limits coupled with few standard limit formulas can handle very complicated expressions and the use of LHR can be minimized to cases where it is really necessary.
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