Prove $\lim_{x\to 2}4x^2-1 =15$ . 
Prove 
  $$
\lim_{x \to 2}\mathrm{f}\left(x\right) =
\lim_{x \to 2}\left(4x^{2} - 1\right) =15
$$

I am getting stuck about the procedure of things. As $x$ gets infinitely closer to $2$, I can see our difference getting close and closer to $0$.
This is what I have so far.
Let $\epsilon> 0$.
Scratch work
$$
\left\vert 4x^{2} - 1 - 15\right\vert =
4\left\vert\left(x - 2\right)\left(x + 2\right)\right\vert
$$
I know I just have the statement if $\ \left\vert x - 2\right\vert < \delta$, then$\ldots$
 A: The key is to first bound $x$ around $2$. 
To that end, we first take $|x-2|<1$ so that $1<x<3$.  
Then, since $3<x+2<5$, we have
$$\begin{align}
|(4x^2-1)-15|&=4|x-2||x+2|\\\\
&\le 20|x-2|\\\\
&<\epsilon
\end{align}$$
whenever $0<|x-2|<\delta=\min\left(1,\frac{\epsilon}{20}\right)$.  And we are done!
A: Since I like having
things approach zero,
let
$x = 2+y$,
so
$x \to 2$
is the same as
$y \to 0$.
Then
$\begin{array}\\
4x^2-1 -15
&=4(2+y)^2-16\\
&=4(4+4y+y^2)-16\\
&=16+16y+4y^2-16\\
&=16y+4y^2\\
&=4y(4+y)\\
& \to 0 \text{ as } y \to 0
\end{array}
$
A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
 \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\half}{{1 \over 2}}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\iff}{\Leftrightarrow}
 \newcommand{\imp}{\Longrightarrow}
 \newcommand{\ol}[1]{\overline{#1}}
 \newcommand{\pars}[1]{\left(\, #1 \,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\, #2 \,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$
$$
\lim_{x \to 2}x = 2\quad\imp\quad
\forall\epsilon' > 0,\ \exists\ \delta' > 0\quad |\quad 
0 < \verts{x - 2} < \delta'\quad\imp\quad\verts{x - 2} < \epsilon'
$$
In particular, any $\epsilon' > 0$ could be written as
$\ds{\epsilon' \equiv {\root{\epsilon + 16} - 4\over 2}}$ for some
$\epsilon > 0$.

\begin{align}
\color{#f00}{\verts{\pars{4x^{2} - 1} - 15}} & =
4\verts{\pars{x - 2}\pars{x + 2}} =
4\verts{\pars{x - 2}^{2} + 4\pars{x - 2}} \leq
4\pars{x - 2}^{2} + 16\verts{x - 2}
\\[3mm] & < 4\epsilon'^{2} + 16\epsilon' = \epsilon
\end{align}

$$
\mbox{Then,}\ \forall\epsilon > 0\,,\quad 
0 < \verts{x - 2} < {\root{\epsilon + 16} - 4\over 2}\quad\imp\quad
\color{#f00}{\verts{\pars{4x^{2} - 1} - 15}} < \epsilon
$$
