Epsilon Delta Proof Verification for $14 + \frac{12}{x^2}$ Can someone check if my steps are completely justified:
Question: Prove $\lim\limits_{x \to \infty} 14 + \frac{12}{x^2} = 14$
For all $\epsilon  > 0 \ \exists \ N>0$ such that $|14+\frac{12}{x^2} -14| < \epsilon$ for all $x > N$
Let $N = \frac{2 \sqrt{3}}{\sqrt{\epsilon}}$, then $x > \frac{2 \sqrt{3}}{\sqrt{\epsilon}} \rightarrow x^2 > \frac{4 \times 3}{\epsilon} = \frac{12}{\epsilon} \rightarrow |\frac{1}{x^2}|<\frac{\epsilon}{12} \rightarrow|\frac{12}{x^2}|< \epsilon \rightarrow |14 +\frac{12}{x^2} -14| = |f(x) - L| < \epsilon $ QED
Thanks

Write your proof clearer:

Proof: Let $\varepsilon>0$ and define $N := \sqrt{\frac{12}{\varepsilon}}$. Then, for all $x > N$ we have $x^2 > \frac{12}\varepsilon$, and so
$$\bigg|\Big( 14 + \frac{12}{x^2} \Big) - 14 \bigg| = \frac{12}{x^2} < \varepsilon.$$

