<s>[INST] How do I find $M>0$ for $\lim _{ x\rightarrow \infty }{ \frac { x^{ 2 }+2x-3 }{ x^2-1 }=1 } $ Given the following limit, find such an $M>0$ that for every $x>M$, the expression is $\frac { 1 }{ 3 } $ close to the limit. In other words find $M>0$ such that for every $x>M:\left| f(x)-L \right| <\frac { 1 }{ 3 } $ for the following function:
$$\lim _{ x\rightarrow \infty  }{ \frac { x^{ 2 }+2x-3 }{ x^2-1 }=1  } $$
So, it seems to me that this question is asking for me to prove that:
If $|f(x)-L|<\varepsilon$, then $x>M$
Steps I took:
$$\left| \frac { x^{ 2 }+2x-3 }{ x^{ 2 }-1 } -1 \right| <\frac { 1 }{ 3 } $$
$$\Longrightarrow \left| \frac { (x+3)(x-1) }{ (x-1)(x+1) } -1 \right| <\frac { 1 }{ 3 } $$
$$\Longrightarrow \left| \frac { (x+3) }{ (x+1) } -\frac { (x+1) }{ (x+1) }  \right| <\frac { 1 }{ 3 } $$
$$\Longrightarrow \left| \frac { 2 }{ x+1 }  \right| <\frac { 1 }{ 3 } $$
$$\Longrightarrow \frac { 2 }{ \left| x+1 \right|  } <\frac { 1 }{ 3 } $$
Without loss of generality, assume that $x>-1$
Then, we get: $$\frac { 2 }{ x+1 } <\frac { 1 }{ 3 } $$
$$\Longrightarrow 2<\frac { 1 }{ 3 } (x+1)$$
$$\Longrightarrow 2<\frac { 1 }{ 3 } x+\frac { 1 }{ 3 } $$
$$\Longrightarrow \frac { 1 }{ 3 } x>2-\frac { 1 }{ 3 } $$
$$\Longrightarrow x>5$$
I don't really know what to do once I hit the "Without loss of generality, assume that..." line. I would like to know if I had the right idea from the beginning of what to prove and where I went wrong in my proof (that is, if it is wrong).
 [/INST] In response to @Cherry_Developer's comment, take $M=5$. Then, for $x>M$, we have: 
\begin{align}
\left|f(x)-L\right|=\left| \frac { x^{ 2 }+2x-3 }{ x^{ 2 }-1 } -1 \right| = \left| \frac { 2 }{ x+1 } \right| < \frac{2}{5+1} = \frac{1}{3}.
\end{align}
</s>