# 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?

## 2 Answers

As $$x\to +\infty$$, it should be $$4x^3\left(e^{\color{blue}{\frac{-2}{x^3}}}-1\right)=4x^3\sum_{k=1}^{\infty}\frac{(\color{blue}{\frac{-2}{x^3}})^k}{k!}= 4x^3\left(-\frac{2}{x^3}+o(1/x^3)\right)=-8+o(1).$$ Then what is the limit?

• Thanks! I see what I did wrong now. Rather than substitute $x$ for $-\frac{2}{x^3}$, I raised it as if my new $x$ value were an exponent. It's always a simple mistake that just derails the entire process. – Curtice Gough Nov 21 '20 at 17:33
• @CurticeGough You are welcome! Making mistakes is a normal part of learning ;-) – Robert Z Nov 21 '20 at 17:48

$$e^{-\dfrac2{x^3}}=1-\dfrac2{x^3}+\dfrac{\left(-\dfrac2{x^3}\right)^3}{2!}+\cdots=1-\dfrac2{x^3}+O\left(\dfrac1{x^6}\right)$$

Alternatively, Set $$-\dfrac2{x^3}=h$$ to find

$$=-4\cdot2\lim_{h\to0^-}\dfrac{e^h-1}h$$