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!

When $M$ is large, it is enough to take $\delta=\frac{1}{2M}$. This is because if we have $\delta=\frac{1}{2M}$ then the numerator is larger than $1$ while the denominator is less than $\left ( 1 + \frac{1}{2M} \right )\frac{1}{2M}$. Hence the quotient is larger than $\frac{2M}{1+\frac{1}{2M}}$. To make this last expression larger than $M$, we need $\frac{1}{2M} \leq 1$, or in other words $M \geq \frac{1}{2}$. 
The problem is that we're in control of $\delta$, not $M$, so we can't require $M \geq \frac{1}{2}$. A workaround is to take $\delta = \min \left \{ \frac{1}{2M},1 \right \}$. Then if $M<\frac{1}{2}$ then the numerator is still larger than $1$ and the denominator is still less than 2, so the quotient is still larger than $\frac{1}{2}>M$.
