Prove $\lim_{x \to 0}$ $\sqrt{3x^2 + 4} = 2$ using the definition We have $\lim_{x \to 0}$ $\sqrt{3x^2 + 4} = 2$
Proof: $\vert \sqrt{3x^2+4} - 2 \vert = \cdots$ I got $\left \vert \frac{3x^2}{\sqrt{3x^2+4}+2} \right \vert = \frac{\vert x \vert \vert 3x \vert}{\sqrt{3x^2+4}+2}$
Then what I do next?
$\vert 3x \vert$ $\lt 6$ ?
 A: You found
$$\vert \sqrt{3x^2+4} - 2 \vert = \left| \frac{3x^2}{\sqrt{3x^2+4}+2} \right|. $$
Hence,
$$\vert \sqrt{3x^2+4} - 2 \vert = \left| \frac{3x^2}{\sqrt{3x^2+4}+2} \right| \leqslant \frac{3}{4}|x|^2$$
and $\vert \sqrt{3x^2+4} - 2 \vert < \epsilon$ if $|x| < \delta = 2\sqrt{\epsilon/3}$.
A: Let $f(x) = \sqrt{3x^2+4}$. Take as a thesis definition.

$\lim_{x \to 0} f(x) = 2$ if and only if for every infinity sequence $\langle a_n\rangle_{\mathbb{N}}$, such that $\lim_{n\to\infty}a_n=0$.
  $$\lim_{x\to 0}f(x)=2 \Leftrightarrow(\forall\epsilon\in\mathbb{R}^+)(\exists\delta\in\mathbb{R})(\forall n\in\mathbb{N})(n\geq\delta\Longrightarrow 2-\epsilon \leq f(a_n) \leq 2+\epsilon)$$

Transform thesis and recive $2 - \epsilon \leq \sqrt{3x^2+4} \leq 2+\epsilon \Leftrightarrow \left(- \epsilon \leq \sqrt{3x^2+4}-2\right) \wedge
\left(\sqrt{3x^2+4}\leq\epsilon + 2\right)$
As $x^2 \nleq 0 \Rightarrow 3x^2+4 > 4 \Rightarrow \sqrt{3x^2+4}-2 >0$, now this and $\epsilon>0 \Rightarrow -\epsilon \leq \sqrt{3x^2+4}-2$. Now you have to prove second part.
As both sides are positive, we can raise to the square both sides of $f(a_n) \leq 2+\epsilon$.
$$\begin{split}
3x^2+4\leq4+4\epsilon + \epsilon^2 \Longleftrightarrow x^2 &\leq\frac{4\epsilon+\epsilon^2}{3} &\overset{\epsilon>0}{\Longleftrightarrow}\\
|x| &\leq \sqrt{\frac{4\epsilon+\epsilon^2}{3}}
\end{split}$$
By definition of limit of sequence, for every $\langle a_n\rangle_{\mathbb{N}}$, such as described in thesis:
$$\left(\lim_{x\to\infty}a_n=0\right)\Longrightarrow\left(\exists \delta \in \mathbb{R}\right)\left(\forall n_{\geq\delta}\in\mathbb{N}\right)\left(|a_n|\leq \sqrt{\frac{4\epsilon+\epsilon^2}{3}}\right)$$
$\mathscr{Q.E.D.}$
