Using exclusively the definition of limit proof that $\lim_{x \to 0} \frac{x^3-2x+x}{\sin(x)} = -1$ Using exclusively the definition of limit proof that 
$$
\lim_{x \to 0} \frac{x^3-x}{\sin(x)} = -1
$$
I have to learn how to prove limits by the delta-epsilon definition, I know how to do basic ones (like linear functions) but not things like this.
I'd like to thoroughly understand the process to get the proof working.
For example: I've seen people choose an "auxiliary $\delta_1=1$" and then declare $\delta=\min(\delta_1,\text{some other thing})$. I have no idea why choosing a delta if allowed, or why we have to take the min when we are done.
Every bit of insight is hugely appreciated.
 A: $$\begin{align}
\left|\frac{x(x^2-1)}{\sin x}+1\right|&=\left|(x^2-1)\frac{x}{\sin x}+1\right|\\\\
&=\left|(x^2-1)\left(\frac{x}{\sin x}-1+1\right)+1\right| \dots \,\,\text{add and subtract}\, 1 \,\text{in the parentheses}  \\\\
&=\left|(x^2-1)\left(\frac{x}{\sin x}-1\right)+x^2\right| \dots \,\,\,\,\text{rearrange terms}  \\\\ 
&\le |x^2-1|\left|\frac{x}{\sin x}-1\right|+x^2 \dots \,\,\,\,\text{apply the triangle inequality} 
\end{align}$$
Given $\epsilon>0$ we know that there exists a $\delta' >0$ such that 
$$\left|\frac{x}{\sin x}-1\right|< \epsilon$$
whenever $0<|x|<\delta'$.  
So, we first choose a number $0<\delta_1 \le \frac12$ so that $|x^2-1|\le \frac34$ whenever $0<|x|<\delta_1$.  
Next, we choose $0<\delta_2  <\sqrt{\epsilon}/2$ so that $x^2<\epsilon/4$ whenever $0<|x|<\delta_2$.  
Thus, if we choose $\delta =\min(1/2,\sqrt{\epsilon}/2,\delta')$ we must have
$$\begin{align}
\left|\frac{x(x^2-1)}{\sin x}+1\right|&=\le |x^2-1|\left|\frac{x}{\sin x}-1\right|+x^2\\\\
&< \frac34 \epsilon + \frac14 \epsilon\\\\
&=\epsilon
\end{align}$$
