Prove that when x approaches to 1-, the function appraochs to negative infinity Prove that 
$$\lim_{x\rightarrow1-}f(x):=\lim_{x\rightarrow1-}\frac{x+2}{2x^2-3x+1}=-\infty.$$
Using the basic definition.
This is the proof from the textbook:
Let $M\in R$ and assume M<0 without losing generality. As x converges to 1 from the left-hand side, $2x^2-3x+1$ is negative and approaches to 0 as x approaches to 1. 
By observing that x has roots 1/2 and 1 and is a parabola opening upward, choosing $\delta\in(0,1)$ such that $1-\delta<x<1$ implies $2/M<2x^2-3x+1<0$;
That is, $-1/(2x^2-3x+1)>-M/2>0$. Since $0<x<1$ also implies $2<x+2<3$, it follows that $-(x+2)/(2x^2-3x+1)>-M$ ; that is,
$$f(x)=\frac{x+2}{2x^2-3x+1}<M$$
for all $1-\delta<x<1$.
I can follow the proof step by step but I have no idea why this question is proved in this way, the "choosing $\delta\in(0,1) $ such that... implies $2/M<2x^2-3x+1<0$" seems jump from no where. Could someone please tell me what's the reasoning behind each step? That is how poeople come up with the idea to solve the problem this way?

I like to let variables
go to zero.
So, in
$\lim_{x\rightarrow1-}\frac{x+2}{2x^2-3x+1}
$
I would let
$x = 1-y$.
Then
$x+2 = 1-y+2
=3-y$
and
$\begin{array}\\
2x^2-3x+1
&=2(1-y)^2-3(1-y)+1\\
&=2(1-2y+y^2)-3+3y+1\\
&=2-4y+2y^2-3+3y+1\\
&=-y+2y^2\\
&=y(-1+2y)\\
\end{array}
$
so
$\lim_{x\rightarrow1-}\frac{x+2}{2x^2-3x+1}
=\lim_{y\rightarrow 0}\frac{3-y}{y(-1+2y)}
$
and it is easy to see why this diverges.
Note that
$\lim_{y\rightarrow 0}\frac{3-y}{-1+2y}
=-3
$.
