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! 
 A: Hint: Write the limit as: 
$\displaystyle\lim_{x \to 0^+}\dfrac{\tan^{-1}(e^x+\tan^{-1}x)-\tan^{-1}(e^{\sin x}+\tan^{-1}\sin x)}{x - \sin x} \cdot \dfrac{x-\sin x}{x^3}$
$= \displaystyle\lim_{x \to 0^+}\dfrac{\tan^{-1}(e^x+\tan^{-1}x)-\tan^{-1}(e^{\sin x}+\tan^{-1}\sin x)}{x - \sin x} \cdot \lim_{x \to 0^+}\dfrac{x-\sin x}{x^3}$
Use the mean value theorem for the first limit and L'Hopital's Rule for the second.
EDIT: Let me elaborate on the first limit. Set $f(x) = \tan^{-1}(e^x + \tan^{-1}x)$. 
Then, $\dfrac{f(x)-f(\sin x)}{x-\sin x} = f'(c_x)$ where $\sin x < c_x < x$. Clearly, $c_x$ depends on $x$. 
However using the squeeze theorem, $\displaystyle\lim_{x \to 0}c_x = 0$. Do you see how to continue?
A: 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.
