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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 $$\left|\frac{3}{x^{2}}-3\right| = \frac{3|1-x^{2}|}{|x^{2}|} = \frac{3|x-1||x+1|}{|x^2|} \le \frac{3|x-1|\left(\frac{9}{4}\right)}{\left(\frac{3}{4}\right)^{2}} = 12|x-1| < \varepsilon.$$

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    $\begingroup$ Sure, whatever works. I wasn't crazy about the given answer to the original question because it didn't show how it was arrived at. I was trying to show my earlier comment how you develop such an argument. $\endgroup$ – Simon S Dec 28 '14 at 17:43
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    $\begingroup$ looks fine to me. by definition, given $\epsilon > 0,$ there exists $\delta >0$ such that $\cdots$. you can have several choice of $\delta > 0$ for a given $\epsilon > 0$ and you can choose any of them. $\endgroup$ – Krish Dec 28 '14 at 17:47
  • $\begingroup$ Perfect thanks a lot both of you. $\endgroup$ – user2850514 Dec 28 '14 at 17:53
  • $\begingroup$ This looks good; I think you just want to change $\delta+2=\frac{9}{4}$ to $\delta+2\le\frac{9}{4}$. $\endgroup$ – user84413 Dec 28 '14 at 18:05
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    $\begingroup$ @user2850514: you need to be little more careful about the inequalities. A lot of them are actually "$\leq$", instead of just "$<$". but it didn't matter at the end. but still $\cdots.$ $\endgroup$ – Krish Dec 28 '14 at 18:05

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