1
$\begingroup$

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 . we have $a=1, b=7, x_0=5$ .
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 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?

  • 0
    @MattPressland, you clear my doubt2012-03-16

1 Answers 1

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").