MHB Limits of Functions .... B&S Theorem 4.2.9 .... ....

[Math Amateur]

I am reading "Introduction to Real Analysis" (Fourth Edition) b Robert G Bartle and Donald R Sherbert ...

I am focused on Chapter 4: Limits ...

I need help in fully understanding an aspect of the proof of Theorem 4.2.9 ...Theorem 4.2.9 ... ... reads as follows:https://www.physicsforums.com/attachments/7257In the above text from Bartle and Sherbert we read the following:"... ... Therefore (why?) it follows that if $$x \in A \cap V_{ \delta } (c), x \neq c$$, then $$f(x) \gt \frac{1}{2} L \gt 0$$. ... ... Can someone please explain why/how it is true that if $$x \in A \cap V_{ \delta } (c), x \neq c$$, then $$f(x) \gt \frac{1}{2} L \gt 0$$?
Hope someone can help ...

Peter

 

[Euge]

I take it $V_\delta(c)$ is the open interval $(c - \delta, c + \delta)$. In that case, if $x\in A\cap V_\delta(c)$ and $x\neq c$, then $x\in A$ and $0 < |x - c| < \delta$. Thus $|f(x) - L| < (1/2)L$, or $-(1/2)L < f(x) - L < (1/2)L$. In particular, $f(x) - L > -(1/2)L$, so $f(x) > L - (1/2)L = (1/2)L > 0$, as desired.

 

[Math Amateur]

I take it $V_\delta(c)$ is the open interval $(c - \delta, c + \delta)$. In that case, if $x\in A\cap V_\delta(c)$ and $x\neq c$, then $x\in A$ and $0 < |x - c| < \delta$. Thus $|f(x) - L| < (1/2)L$, or $-(1/2)L < f(x) - L < (1/2)L$. In particular, $f(x) - L > -(1/2)L$, so $f(x) > L - (1/2)L = (1/2)L > 0$, as desired.

Thanks Euge ...

Appreciate your help ...

Peter