<s>[INST] How to find $\frac{0}{0}$ limit without L'Hôpital's rule I am having trouble solving this limit. I tried applying L'Hôpital's rule but I got $\frac{0}{0}$.
$$\lim_{x\to0} {\frac{\frac{1}{1+x^3} + \frac{1}{3}\log{\left(1+3x^3\right)}-1}{2\sin{\left(3x^2\right)}-3\arctan{\left(2x^2\right)}}}$$
I would appreciate any hints in the right direction. Thanks in advance for your help.
 [/INST] Hint: $$\frac{1}{1+x^3} = 1 - x^{3} + x^{6} - x^{9} + O(x^{12}) \,\,\, \text{and} \,\,\, \log (1 + 3x^3) = 3x^3 - \frac{9x^6}{2} - 9x^9 + O(x^{12})$$
$$\sin (3x^2) = 3x^2  - \frac{9x^6}{2} + O(x^{10}) \,\,\, \text{and} \,\,\, \arctan (2x^2) = 2x^2 - \frac{8x^6}{3} + O(x^{10}) $$
Can you take it from here? 
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