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Definition:

Let f(x) be a function defined on an interval that contains x=a, except possibly at x=a . Then we say that,if for every number ε>0 there is some number δ>0 such that

$$\left| {f\left( x \right) - L} \right| < \varepsilon \hspace{0.5in}{\mbox{whenever}}\hspace{0.5in}0 < \left| {x - a} \right| < \delta$$

My question is whether I can change the def to the following
1): $$\left| {f\left( x \right) - L} \right| <= \varepsilon \hspace{0.5in}{\mbox{whenever}}\hspace{0.5in}0 < \left| {x - a} \right| <= \delta$$ 2):
$$\left| {f\left( x \right) - L} \right| < \varepsilon \hspace{0.5in}{\mbox{whenever}}\hspace{0.5in}0 < \left| {x - a} \right| <= \delta$$ 3): $$\left| {f\left( x \right) - L} \right| <= \varepsilon \hspace{0.5in}{\mbox{whenever}}\hspace{0.5in}0 < \left| {x - a} \right| < \delta$$

If not, could you please provide an counter-example? It seems all ok to me. But I am not certain.

AlexWei
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    They are all equivalent. – Kavi Rama Murthy Mar 27 '21 at 06:13
  • ^ and it's a good exercise for you to see why they're all equivalent – peek-a-boo Mar 27 '21 at 06:16
  • @peek-a-boo I will try to prove myself. thanks. I am watching khanacademy calculus. The question always popup in my head over and over. : ) – AlexWei Mar 27 '21 at 06:19
  • Hint: If you know that something is $\leq\varepsilon$ for all $\varepsilon>0$, but then you want $<\varepsilon$ instead, just think about the fact that this holds for all $\varepsilon > 0$ -- what can you do to $\delta$ to change that $\leq$ into $<$? – Brian Mar 27 '21 at 06:27
  • @HansLundmark Yes. Totally. Thanks a lot. My question is the same as this one. – AlexWei Mar 27 '21 at 07:50

1 Answers1

1

Given $$\left| {f\left( x \right) - L} \right| <= \varepsilon \hspace{0.5in}{\mbox{whenever}}\hspace{0.5in}0 < \left| {x - a} \right| <= \delta$$

According to this def, since $\varepsilon$ is arbitrary small. Given $\varepsilon$ we can find a $\varepsilon_1$<$\varepsilon$ which also meet the condition. In this case: $$\left| {f\left( x \right) - L} \right| < \varepsilon_1 \hspace{0.5in}{\mbox{whenever}}\hspace{0.5in}0 < \left| {x - a} \right| <= \delta_1$$

When $0 < \left| {x - a} \right| <= \delta_1$, the $\left| {f\left( x \right) - L} \right| < \varepsilon_1$ will always fullfill. So when pick a number $\delta_2$<$\delta_1$,$\left| {f\left( x \right) - L} \right| < \varepsilon_1$ still holds.

Thus we have: $$\left| {f\left( x \right) - L} \right| < \varepsilon_1 \hspace{0.5in}{\mbox{whenever}}\hspace{0.5in}0 < \left| {x - a} \right| < \delta_2$$

This could also be represented as $$\left| {f\left( x \right) - L} \right| < \varepsilon \hspace{0.5in}{\mbox{whenever}}\hspace{0.5in}0 < \left| {x - a} \right| < \delta$$

AlexWei
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