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!

In general, for any $A,\ B\neq0,\ C\neq0,$
$$\frac{A}{\left(\frac 1B\right)C} = \frac{1}{\left(\frac 1B\right)} \cdot \frac AC
 = B \cdot \frac{A}{C}.$$
Therefore
$$\frac{0.5}{\left(\frac{1}{5M}\right)(1.5+1)} = 5M \cdot \frac{0.5}{1.5+1}
 = 5M \cdot 0.2.$$
(Assuming the left-hand end of the formula above contains the "complicated terms" that we had to "get rid of".)
