I am writing a question referring to another question Formal proof of limit. The accepted solution posed to this question takes $\delta<\min{\left(\frac{1}{2},\frac{\varepsilon}{30}\right)}$ but could we in fact take $\delta=\min{\left(\frac{1}{4}, \frac{\varepsilon}{12}\right)}$ and anything similar?
Suppose we are given $\varepsilon>0$ then we choose $\delta=\min{\left(\frac{1}{4}, \frac{\varepsilon}{12}\right)}$, then $$|x-1|=|1-x|\ge|1|-|x| = 1 - |x| \implies |x|\ge\frac{3}{4}.$$ Also $$|x+1| = |x-1+2|\le|x-1|+2<\delta+2\le\frac{9}{4}.$$ So $$\left|\frac{3}{x^{2}}-3\right| = \frac{3|1-x^{2}|}{|x^{2}|} = \frac{3|x-1||x+1|}{|x^2|} \le \frac{3|x-1|\left(\frac{9}{4}\right)}{\left(\frac{3}{4}\right)^{2}} = 12|x-1| < \varepsilon.$$