<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] Recall some standard limits (which can be computed without using
l'Hospital's rule ):
\begin{eqnarray*}
\lim_{u\rightarrow 0}\frac{\frac{1}{1+u}-1+u}{u^{2}} &=&1 \\
\lim_{u\rightarrow 0}\frac{\log (1+u)-u}{u^{2}} &=&-\frac{1}{2} \\
\lim_{u\rightarrow 0}\frac{\sin u-u}{u^{3}} &=&-\frac{1}{6} \\
\lim_{u\rightarrow 0}\frac{\arctan u-u}{u^{3}} &=&-\frac{1}{3}.
\end{eqnarray*}
Now one can re-write the original expression using those of the standard
limits above as follows 
\begin{eqnarray*}
{\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) }}} &{=}&\frac{\left( 
\frac{1}{1+x^{3}}-1+x^{3}\right) +\left( \frac{1}{3}\log {\left(
1+3x^{3}\right) -x}^{3}\right) }{\left( 2\sin {\left( 3x^{2}\right) -6x}%
^{2}\right) -\left( 3\arctan {\left( 2x^{2}\right) -6x}^{2}\right) } \\
&=&\frac{\left( \frac{\frac{1}{1+(x^{3})}-1+(x^{3})}{(x^{3})^{2}}\right)
+3\times \left( \frac{\left( \log {\left( 1+\left( 3x^{3}\right) \right) -}%
\left( {3x}^{3}\right) \right) }{(3x^{3})^{2}}\right) }{2\times 3^{3}\left( 
\frac{\sin {\left( 3x^{2}\right) -(3x}^{2})}{(3x^{2})^{3}}\right) -3\times
2^{3}\left( \frac{\arctan {\left( 2x^{2}\right) -(2x}^{2})}{\left(
2x^{2}\right) ^{3}}\right) }
\end{eqnarray*}
It follows that 
\begin{equation*}
\lim_{x\rightarrow 0}\left( \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) }}\right) =\frac{\left( 1\right) +3\times \left( -\frac{1}{2}%
\right) }{2\times 3^{3}\left( -\frac{1}{6}\right) -3\times 2^{3}\left( -%
\frac{1}{3}\right) }=\frac{1}{2}.
\end{equation*}
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