<s>[INST] how to evaulate: $\lim \limits_{x \to 0} \frac{\ln(1+x^5)}{(e^{x^3}-1)\sin(x^2)} $ How do I evaluate: $\lim \limits_{x \to 0} \frac{\ln(1+x^5)}{(e^{x^3}-1)\sin(x^2)} $ ?
according to Taylor's series, I did like this: 
$$\lim \limits_{x \to 0} \frac{\ln(1+x^5)}{(e^{x^3}-1)\sin(x^2)}=\lim \limits_{x \to 0} \frac{x^5 - \frac{x^{10}}{2}+ O_3(x)}{(x^3+ \frac{x^6}{2!} +O_4(x))(x^2- \frac{x^6}{3!} +\frac{x^{10}}{5!} +O_4(x))} $$
but how do I continue from here? 
 [/INST] We have 
\begin{align}
&\lim_{x\to 0} \frac{x^5 - \frac{x^{10}}{2} + O(x^{15})}{(x^3 + \frac{x^6}{2!} + O(x^9))(x^2 - \frac{x^6}{3!} + O(x^{10}))}\\
&= \lim_{x\to 0}\frac{1 + O(x^5)}{(1 + O(x^3))(1 + O(x))}\\\\
&= \frac{1}{(1)(1)}\\
&= 1.
\end{align}
</s>