algebra precalculus - Evaluate $lim_{x to 0} frac{tan(tan x)-sin(sin x)}{tan x-sin x}$

Evaluate

$$\lim_{x \to 0}\frac{\tan(\tan x)-\sin(\sin x)}{\tan x-\sin x}$$

First I tried using L'Hopital's rule..but it's very lengthy

Next I have written the limits as

$$L=\lim_{x \to 0}\frac{\tan(\tan x)-\sin(\sin x)}{\tan x-\sin x}=\frac{\lim_{x\to 0}\frac{\tan(\tan x)-\sin(\sin x)}{x^3}}{\lim_{x \to 0}\frac{\tan x-\sin x}{x^3}}=\frac{L_1}{L_2}$$

Now by L'Hopital's Rule we get $L_2=0.5$

$$L_1=\lim_{x\to 0}\frac{\tan(\tan x)-\sin(\sin x)}{x^3}$$

Now $L_1$ can also be evaluated using three applications of L'Hopital's Rule, but is there any other approach?