Limit of $\mu(\theta)=\frac{e^{\theta}(\theta-1)+1}{(e^{\theta}-1)\theta}$ when $\theta \to \pm \infty$ & $\theta \to 0$

Question

Let $\mu(\theta)$ be defined as:

$$\mu(\theta)=\frac{e^{\theta}(\theta-1)+1}{(e^{\theta}-1)\theta}.$$

I would like to show limit when $\theta \to \pm \infty$, I know that it is $0$ and $1$, but I would like to come to this result.

$$\frac{e^{\theta}(\theta-1)+1}{(e^{\theta}-1)\theta}=\frac{\left ( 1+\theta+\frac{\theta^2}{2}+\frac{\theta^3}{6}+\mathcal{O}(\theta^4) \right ) \left ( 1-\frac{1}{\theta} \right )+\frac{1}{\theta}}{1+\theta+\frac{\theta^2}{2}+\frac{\theta^3}{6}+\mathcal{O}(\theta^4)-1}=\frac{\frac{\theta}{2}+\frac{\theta^2}{3}+\frac{\theta^3}{6}+\mathcal{O}(\theta^4)-\mathcal{O}(\theta^3)}{\theta+\frac{\theta^2}{2}+\frac{\theta^3}{6}+\mathcal{O}(\theta^4)}$$ Now I don't know how to justify when $\theta \to -\infty$ is $0$ and $\theta \to \infty$ is $1$.

EDIT: After previous question is solved, I need to show also when $\theta \to 0$ then $\mu(\theta) \to 1/2$.

Answer 1

Divide numerator and denominator by $\theta e^{\theta}$ to obtain $$ \frac{1 - \theta^{-1} + \theta^{-1}e^{-\theta}}{1 - e^{-\theta}}. $$ We have $\lim_{\theta\rightarrow\infty}e^{-\theta} = 0$ and $\lim_{\theta\rightarrow\infty}\theta^{-1}= 0$, so $$ \lim_{\theta\rightarrow\infty}\frac{1 - \theta^{-1} + \theta^{-1}e^{-\theta}}{1 - e^{-\theta}} = \frac{1-0+0}{1-0} = 1. $$ Can you do $\theta\rightarrow-\infty$ yourself?

Answer 2

As an alternative note that

$$\frac{e^{\theta}(\theta-1)+1}{(e^{\theta}-1)\theta}=\frac{\theta e^{\theta}-\theta +\theta -e^{\theta}+1 }{(e^{\theta}-1)\theta}=1+\frac{1 }{e^{\theta}-1}-\frac1 \theta$$