Evaluating limits using a series I'm trying to use a Taylor series centered at $0$ to evaluate this limit:
$$\lim_{x\to \infty}4x^3(e^\frac{-2}{x^3}-1)$$
I rewrote the function as its Maclaurin series:
$$4x^3(e^\frac{-2}{x^3}-1)=\sum_{k=1}^\infty\frac{4x^{3-\frac{2k}{x^3}}}{{k!}}$$
In expanded form:
$$\sum_{k=1}^\infty\frac{4x^{3-\frac{2k}{x^3}}}{{k!}}=4x^{3-\frac{2}{x^3}}+\frac{4x^{3-\frac{4}{x^3}}}{2!}+\frac{4x^{3-\frac{6}{x^3}}}{3!}+...$$
As $x$ goes to $\infty$, $\frac{2}{x^3}$ goes to $0$.  Thus, the limit of the first term is simply the limit of $4x^3$, which is $\infty$.  Based on this fact alone, I would assume, the limit of the entire series is $\infty$, but apparently the answer is $-8$.  What did I do wrong?

$$e^{-\dfrac2{x^3}}=1-\dfrac2{x^3}+\dfrac{\left(-\dfrac2{x^3}\right)^3}{2!}+\cdots=1-\dfrac2{x^3}+O\left(\dfrac1{x^6}\right)$$
Alternatively, Set $-\dfrac2{x^3}=h$ to find
$$=-4\cdot2\lim_{h\to0^-}\dfrac{e^h-1}h$$
