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? 

From Taylor's expansions at $x=0$, $\log(1+x^5) \approx x^5$, $e^{x^3}-1 \approx x^3$ and $\sin x^2 \approx x^2$, as $x \rightarrow 0$. Thus,
$$\lim \limits_{x \rightarrow 0} \frac{\log(1+x^5)}{(e^{x^3}-1)(\sin x^2 )} = \lim \limits_{x \rightarrow 0} \frac{x^5}{x^{3} x^{2}} =1.$$
