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Following question and answer are from Thomas calculus book:

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

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! – checkmath Mar 16 '12 at 11:28
Check the order of the implication - you want $\delta$ so that if $0<|x-x_0|<\delta$ then $a<x<b$, which you have. Your example shows that it is not true that $a<x<b\implies0<|x-x_0|<\delta$, but that's not a problem. – Matthew Pressland Mar 16 '12 at 11:33
@MattPressland, you clear my doubt – Vikram Mar 16 '12 at 11:43

In the case $x = 1$, the statement $0 < \lvert x - x_0 \rvert < \delta$ is false. Therefore, the implication $0 < \lvert x - x_0 \rvert < \delta \implies a < x < b$ is true (as it has the form "false $\implies$ false").

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