Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $a=1, b=7, x_0=5$ .
Solution:
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?

share|improve this question
    
That is the greatest delta satisfying such property, aka optimum delta! –  checkmath Mar 16 '12 at 11:28
1  
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. –  Matt Pressland Mar 16 '12 at 11:33
    
@MattPressland, you clear my doubt –  Vikram Mar 16 '12 at 11:43

1 Answer 1

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").

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.