Use the definition of infinite limit to prove $\lim_{x \to 1+} \frac{x}{x^2-1}=\infty$ 
Prove
  $$\lim_{x \to 1+} \frac{x}{x^2-1}=\infty$$

And I was given the solution like this: but I could not understand how it removes the complicated terms.
Let $\delta=\min(0.5,\frac{1}{5M})$.  
$$\frac{x}{x^2-1}=\frac{x}{(x+1)(x-1)}
\geq\frac{0.5}{\left(\frac{1}{5M}\right)(1.5+1)}=5M\times0.2=M$$
I understand the definition of $M-\delta$, but what I don't understand is what is the solution doing, I mean the process of estimation to get rid of the complicated terms. Anyone can enlighten me? thanks!

We can first use partial fraction decomposition to $\frac{x}{x^2-1},$ which leads to
\begin{gather*}
\frac{x}{x^2-1}=\frac{1}{2(x+1)}+\frac{1}{2(x-1)},
\end{gather*}
which implies that 
\begin{gather*}\tag{1}
\frac{x}{x^2-1}>\frac{1}{2(x-1)}, \qquad \text{provided } x>-1.
\end{gather*}
Let $M>0.$ Set 
$$\delta=\frac{1}{2M}.$$
For every $x>1,$ we have
\begin{align*}
0<x-1<\delta\implies  \frac{x}{x^2-1}&>\frac{1}{2(x-1)}\qquad\qquad \text{by }(1)\\
&>\frac{1}{2\delta}=\frac{1}{2\frac{1}{2M}}=M,
\end{align*}
which, by definition of limit, completes the proof.
