<s>[INST] 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?
$\vert 3x \vert$ $\lt 6$ ?
 [/INST] 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 $|x| < \delta = 2\sqrt{\epsilon/3}$.
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