How do I complete the proof of proving the $\lim _{ x\rightarrow \infty }{ \frac { x^{ 2 }+1 }{ (x+1)^{ 2 } } =1 } $ $$\lim _{ x\rightarrow \infty  }{ \frac { x^{ 2 }+1 }{ (x+1)^{ 2 } } =1 } $$
Proof: Let $\epsilon > 0$
Then, $$ \left| \frac { x^{ 2 }+1 }{ (x+1)^{ 2 } } -1 \right| <\epsilon $$
$$\Longleftrightarrow \left| \frac { x^{ 2 }+1 }{ (x+1)^{ 2 } } -\frac { (x+1)^{ 2 } }{ (x+1)^{ 2 } }  \right| <\epsilon $$
$$\Longleftrightarrow \left| \frac { x^{ 2 }+1 }{ (x+1)^{ 2 } } -\frac { x^{ 2 }+2x+1 }{ (x+1)^{ 2 } }  \right| <\epsilon $$
$$\Longleftrightarrow \left| \frac { -2x }{ (x+1)^{ 2 } }  \right| <\epsilon $$
$$\Longleftrightarrow \left| \frac { -2 }{ x^{ 3 }+2x^{ 2 }+x }  \right| <\epsilon $$
$$\Longleftrightarrow \frac { 2 }{ \left| x^{ 3 }+2x^{ 2 }+x \right|  } <\epsilon $$
Now, we will calculate the lower bound on $x$ to force $ \left| \frac { x^{ 2 }+1 }{ (x+1)^{ 2 } } -1 \right| <\epsilon $ to hold true.
Without loss of generality, assume that $x>0$, then we get: $$\frac { 2 }{ x^{ 3 }+2x^{ 2 }+x } <\epsilon $$
$$\Longrightarrow 2<\epsilon (x^{ 3 }+2x^{ 2 }+x)$$
$$\Longrightarrow 2<\epsilon x^{ 3 }+\epsilon 2x^{ 2 }+\epsilon x$$
At this point I get stuck. How do I go from here? I can't find a way in which I could manipulate this to get it in the form $x>...$ A hint in the right direction would be beneficial. 
 A: Given $\epsilon > 0$ we want to find an $M$ such that $x > M \Rightarrow | \cdots | < \epsilon$. 
Remember you just need to find an $M$; make life as easy for yourself as you can and don't care about finding a low $M$.
Hence, I would simplify and write for all $x > 0$,
$$ \left| \frac{x^2+1}{(x+1)^2} -1 \right|  = \left| \frac { -2x }{ (x+1)^{ 2 } }  \right| \leq \frac{2x}{x^2} = \frac{2}{x}$$
Thus given $\epsilon > 0$, choose $M = 2/\epsilon$ (which is positive).
Then 
$$x > M \ \Rightarrow \ x > \frac 2\epsilon \ \Rightarrow \ \frac 2x < \epsilon \ \Rightarrow \  \left| \frac{x^2+1}{(x+1)^2} -1 \right|  < \frac 2x <\epsilon$$

Added (answer to question in comments):
We have for all positive $x$,
$$\left| \frac{x^2+1}{(x+1)^2} -1 \right| < \frac 2x .$$
Hence given $\epsilon > 0$ if we can find an $M > 0$ such that
$$x > M \quad \Rightarrow \quad \frac 2x < \epsilon$$
this implies
$$x > M \quad \Rightarrow \quad \left| \frac{x^2+1}{(x+1)^2} -1 \right|  < \epsilon .$$
Now, bounding $\frac 2x$ by $\epsilon$ is straight forward:
$$\frac 2x < \epsilon \quad\text{ iff }\quad  x > \frac 2\epsilon$$
Hence the choice of $M = 2/\epsilon$ does the trick.
A: Rewrite it as
$$
\lim_{x\to\infty}\frac{1+\dfrac{1}{x^2}}{1+\dfrac{2}{x}+\dfrac{1}{x^2}}
=1
$$
For $x>0$, the denominator is $>1$ and it's not restrictive to assume it.
The inequality to be solved becomes then
$$
\left|-\frac{2}{x}\right|<
\varepsilon\left(1+\dfrac{2}{x}+\dfrac{1}{x^2}\right)
$$
which is certainly satisfied if
$$
\frac{2}{x}<\varepsilon
$$
so for
$$
x>\frac{2}{\varepsilon}
$$
