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!
 A: 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 $\frac{1}{2}>M$.
A: 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".)
A: 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.
A: 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.
