From the uploaded image, I fail to understand why the correct (or permissible) value for delta is the minimum among the two values. Why or how is the minimum value the correct one, why not the other way around? Thanks.
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2what do you mean by "the other way around"? – Yanko Aug 10 '17 at 18:17
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See my answer here: https://math.stackexchange.com/questions/2354988/proving-continuity-of-fx-x2/2355006#2355006 – Sean Roberson Aug 10 '17 at 18:22
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You need to find one value of $\delta$ that forces $|x^2 - 4|$ to be small enough. There are some cases in which $2 - \sqrt{4-\varepsilon}$ is required, and other cases in which $\sqrt{4+\varepsilon} - 2$ is required. If $|x-2|$ is smaller than only one of those, we might not get what we want; but if it is smaller than both, then we are in business. Hence we take the minimum. – Xander Henderson Aug 10 '17 at 18:22
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First of all the book's solution is wrong (because of a circular argument which is difficult to detect). Next the way it has been presented is worse. The same symbol $\delta$ has been used to denote two different quantities. Also such a solution is bound to confuse readers and your question is a perfect outcome of studying this solution. I refer you to a similar example for more information: https://math.stackexchange.com/a/2349196/72031 Another answer (https://math.stackexchange.com/a/1333329/72031) may also be helpful. – Paramanand Singh Aug 11 '17 at 02:53