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Following question and answer are from Thomas calculus book:

Find a value of $\delta >0$ such that for all $0< |x-x_0|< \delta \implies a Solution:
Step 1: $|x-5|<\delta \implies -\delta< x-5< \delta \implies -\delta+5 Step 2: $ \delta+5=7 \implies \delta=2,$ or $-\delta+5=1 \implies \delta=4$ The value of $\delta$ which assures $|x-5|< \delta \implies 1

My question:when we consider $x=1, |x-5|=|1-5|=4$, and it is not less than $\delta=2$, then how come we take $\delta$ to be equal to $2$? where am I going wrong?

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    That is the greatest delta satisfying such property, aka optimum delta!2012-03-16
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    Check the order of the implication - you want $\delta$ so that if $0<|x-x_0|<\delta$ then $a, which you have. Your example shows that it is not true that $a, but that's not a problem.2012-03-16
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    @MattPressland, you clear my doubt2012-03-16

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