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!

Given an $M>0$ somebody found by reverse engineering the proposed $\delta:=\min\left\{0.5,{1\over 5 M}\right\}$ would do the job. Now insert after this definition of $\delta$ the phrase "If $1<x<1+\delta$"; saying that you are only considering points $x$ with  $|x-1|<\delta$ and $x>1$. Then your text could read as follows:
If $1<x<1+\delta$ then $0<x-1<{1\over 5M}$, and $x+1<1.5$. It follows that
$${x\over x^2-1}={x\over(x-1)(x+1)}>{1\over{1\over 5M}\cdot1.5}> M\ .\tag{1}$$
This means that the proposed $\delta$ does exactly what it was set up to do.
Note that, given $M$, such a $\delta>0$ is not determined uniquely, even though people write $\delta(M)$ sometimes. It is experience with such questions that leads us in finding an admissible $\delta$. Here are some hints: The factor $x-1$ in the denominator of $(1)$ indicates that ${1\over\delta}$ should have the order of magnitude at least $M$. Since the other factors amount to ${x\over1+x}<1$ some extra measures have to be taken to guarantee $(1)$. Nobody is claiming that the proposed $\delta$ is the "optimal" or "simplest" $\delta=\delta(M)$ doing the job.
