<s>[INST] find this limit $\lim_{x\to0^{+}}\frac{\tan{(\tan{x})}-\tan{(\sin{x})}}{\tan{x}-\sin{x}}$ find the limit.

$$\lim_{x\to0^{+}}\dfrac{\tan{(\tan{x})}-\tan{(\sin{x})}}{\tan{x}-\sin{x}}$$

my try:
$$\tan{x}=x+\dfrac{1}{3}x^3+o(x^3),\sin{x}=x-\dfrac{1}{6}x^3+o(x^3)$$
so
$$\tan{(\tan{x})}=\tan{x}+\dfrac{1}{3}(\tan{x})^3+o(\tan^3{x})=x+\dfrac{1}{3}x^3+\dfrac{1}{3}(x+\dfrac{1}{3}x^3)^3+o(x^3)$$
$$\tan{(\sin{x})}=\sin{x}+\dfrac{1}{3}(\sin{x})^3+o(\sin^3{x})=x-\dfrac{1}{6}x^3+\dfrac{1}{3}(x-\dfrac{1}{6}x^3)^3+o(x^3)$$
so
$$\tan{(\tan{x})}-\tan{(\sin{x})}=\dfrac{1}{2}x^3+o(x^3)$$
$$\tan{x}-\sin{x}=\dfrac{1}{2}x^3+o(x^3)$$
Have other metods? Thank you 
 [/INST] Factor a $\tan(x)$ from the denominator:
$$\frac{\tan\tan(x) - \tan\sin(x)}{\tan(x)-\sin(x)} = \frac{\frac{\tan\tan(x)}{\tan(x)} - \cos(x)\frac{\tan\sin(x)}{\sin(x)}}{1-\cos(x)}.$$
The quotients in the numerator go to $1$, so the numerator goes to $1-\cos(x)$. This suggests cancellation.
After two derivatives, the denominator of this quotient will go to $1$, so one strategy is L'Hopital's rule. (I don't have time to compute it but maybe I'll have time later today.)
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