Another Fun Limit

Relevant wiki: Limits by Taylor series

Although the limit is of the form $\frac{0}{0}$ , L'Hospital's rule would get rather messy rather quickly. So we use the Taylor expansions of $e^{x^3}$ and $\ln(1+x)$ to get: $\begin{aligned} &\phantom{{}={}} \lim_{x\rightarrow 0} \frac{(e^{x^3} - x^3 - 1)(\ln(1+x) - x)}{x^k} \\ &= \lim_{x\rightarrow 0} \frac{[(1 + \frac{x^3}{1!} + \frac{x^6}{2!} + \frac{x^9}{3!} + \cdots) - x^3 - 1][(x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \cdots) - x]}{x^k} \\ &= \lim_{x\rightarrow 0} \frac{( \frac{x^6}{2} + \frac{x^9}{6} + \cdots)(- \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} +\cdots )}{x^k} \\ &= \lim_{x\rightarrow 0} \frac{( -\frac{x^8}{4} + \frac{x^9}{6} - \frac{x^{10}}{8} +\cdots)}{x^k} \end{aligned}$ Now, we see that if $k > 8$ , the limit will be infinite, and if $k < 8$ , the limit will be zero. Thus the only value of $k$ for which the expression is non-zero and finite is $k = 8$ .