I want you to solve this limit in as many ways as ...

Below are five independent “flavours” of attack on

$$\boxed{\displaystyle\lim_{a\to 0}\; \frac{\tan(\tan a)-\sin(\sin a)}{\tan a-\sin a}} \qquad (=\,2)$$

Each path highlights a different classical idea, so you can choose the style that resonates with you.

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1. Maclaurin-series (brute-force)

Write every ingredient as a power series about 0 (radians):

$$\begin{aligned} \tan a &= a+\frac{a^{3}}{3}+\frac{2a^{5}}{15}+O(a^{7}),\\[2mm] \sin a &= a-\frac{a^{3}}{6}+\frac{a^{5}}{120}+O(a^{7}). \end{aligned}$$

Denominator

$$\tan a-\sin a = \frac{a^{3}}{2}+\frac{a^{5}}{8}+O(a^{7}).$$

Numerator — first set $T=\tan a$ and $S=\sin a$.

$$\begin{aligned} \tan(\tan a)&=\tan T =T+\frac{T^{3}}{3}+\frac{2T^{5}}{15}+O(a^{7}),\\ \sin(\sin a)&=\sin S =S-\frac{S^{3}}{6}+\frac{S^{5}}{120}+O(a^{7}). \end{aligned}$$

Insert $T,S$ from the first line and keep terms up to $a^{5}$:

$$\tan(\tan a)-\sin(\sin a)=a^{3}+ \frac{a^{5}}{2}+O(a^{7}).$$

Hence

$$\frac{\tan(\tan a)-\sin(\sin a)}{\tan a-\sin a} =\frac{a^{3}+a^{5}/2+O(a^{7})}{a^{3}/2+a^{5}/8+O(a^{7})} =\frac{2\bigl(1+\frac{a^{2}}{2}+O(a^{4})\bigr)} {1+\frac{a^{2}}{4}+O(a^{4})}\xrightarrow[a\to0]{}2.$$

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2. Repeated L’Hospital (classical analysis)

Both numerator and denominator vanish to order 3 at $a=0$.
Differentiate until the first non–zero derivative appears.

1st derivative (still zero), 2nd derivative (still zero).
At the 3rd derivative you get

$$\frac{(\tan(\tan a)-\sin(\sin a))^{(3)}} {(\tan a-\sin a)^{(3)}}\Big|_{a=0} =\frac{6}{3}=2.$$

The bookkeeping is long by hand, but symbol-manipulation software or a tidy chain-rule computation confirms those numbers quickly.

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3. “Equivalent infinitesimal” trick

For small $a$:

$$\begin{aligned} \tan a-\sin a &\sim \frac{a^{3}}{2},\\ \tan(\tan a)-\sin(\sin a) &=\bigl[\tan a + O(a^{3})\bigr]-\bigl[\sin a + O(a^{3})\bigr]\\ &\sim (\tan a-\sin a)+\frac{a^{3}}{2} \;\sim\;a^{3}. \end{aligned}$$

So the quotient behaves like

$$\frac{a^{3}}{a^{3}/2}\;\longrightarrow\;2.$$

The beauty is that only the leading‐order infinitesimals are needed; everything else is squeezed into the $O(a^{3})$ remainder.

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4. Cauchy’s Mean-Value Theorem (one-line elegance)

Let

$$F(a)=\tan(\tan a)-\sin(\sin a),\qquad G(a)=\tan a-\sin a .$$

Both vanish at $a=0$. Cauchy’s MVT gives, for $a\neq0$,

$$\frac{F(a)}{G(a)}=\frac{F'(\xi)}{G'(\xi)} \quad\text{for some } \xi\in(0,a).$$

Compute the first derivatives only once:

$$\begin{aligned} F'(x)&=(\sec^{2}(\tan x))\sec^{2}x-\cos(\sin x)\cos x,\\ G'(x)&=\sec^{2}x-\cos x. \end{aligned}$$

Take the limit $x\to0$:

$$F'(x)\sim (1)(1) - (1)(1) + \tfrac43x^{2},\quad G'(x)\sim 1-1 + \tfrac12x^{2},$$

so $F'(\xi)/G'(\xi)\to \dfrac{\tfrac43}{\tfrac12}=2$.
Therefore $F(a)/G(a)\to2$.

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5. Derivatives-of-composites shortcut (Faà di Bruno flavour)

Because $\tan x-\sin x$ is an odd function whose first non-zero derivative at 0 is the third, the limit equals

$$\frac{f^{(3)}(0)}{g^{(3)}(0)},$$

where

$$f(a)=\tan(\tan a)-\sin(\sin a),\quad g(a)=\tan a-\sin a.$$

Use the chain rule (or Faà di Bruno) just for the third derivatives at 0:

$$f^{(3)}(0)=6,\qquad g^{(3)}(0)=3\quad\Longrightarrow\quad \frac{f^{(3)}(0)}{g^{(3)}(0)}=2.$$

This route needs no expansions beyond the exact derivatives at a single point.

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Bottom line

Whatever your philosophical stance—series, differential calculus, asymptotics, or the artful squeeze—the answer harmonises at

$$\boxed{2}.$$

Feel free to pick your favourite melody!