Proving limit properties From Trench Real Analysis







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*What if $\epsilon \ge 1$? How does the proof not hold?

*How do we have $|g(x)| > \frac{|L_2|}{2}$?
 A: Part 1. If you have a $\delta$ corresponding to some $\varepsilon < 1$, then take the same $\delta$ for any $\varepsilon \geqslant 1$. This can simply be seen by choosing $\varepsilon < 1$ and $\eta \geqslant 1$. If $$\left| f(x) - L\right| < \varepsilon$$ whenever $0 < \left| x - x_0\right| < \delta$, then for sure for the same $\delta > 0$ $$\left| f(x) - L\right| < \varepsilon < \eta$$
Part 2. We have $$\left|\left| g(x)\right| - \left|L_2 \right|\right| \leqslant \left| g(x) - L_2 \right| < \frac{\left| L_2\right|}{2}$$ by the reversed triangle inequality and thus $$-\frac{\left| L_2\right|}{2}< \left| g(x)\right| - \left|L_2 \right| < \frac{\left| L_2\right|}{2}$$ Hence by the left side inequality $$-\frac{\left| L_2\right|}{2}< \left| g(x)\right| - \left|L_2 \right| \Leftrightarrow \frac{\left| L_2\right|}{2}< \left| g(x)\right|$$
A: *

*Who cares? When doing analysis, you usually consider $\epsilon>0$ a "small enough" parameter, so you are not interested in knowing what happens if $\epsilon$ is bigger than something.

*Go on reading the proof. You have cut out the explanation from your post.

