Example of limit of a function

Question

I'm self-studying from the book Understanding Analysis by Stephen Abbott, and I don't understand example 4.2.2 on page 105.

The aim of the example is to show that: $$ \lim_{x \to 2} g(x) = 4 $$ where $g(x) = x^2$.

The author starts by writing: $$ |g(x) - 4| = |x^2 - 4 | = |x+2||x-2| $$ which I understand. But then the following paragraph makes no sense to me:

We can make $|x-2|$ as small as we like, but we need an upper bound on $|x+2|$ in order to know how small to choose $\delta$. The presence of the variable $x$ causes some initial confusion, but keep in mind that we are discussing the limit as $x$ approaches $2$. If we agree that our $\delta$-neighborhood around $c = 2$ must have radius no bigger than $\delta = 1$, then we get the upper bound $|x + 2| \leq |3 + 2| = 5$ for all $x \in V_\delta(c)$.

By saying we are only considering the $\delta$-neighborhood around $c=2$ with a radius smaller than or equal to $1$, I would think we get: $$ V_\delta (c) = \{ x \in \mathbb{R} \mid |x-2| < 1 \} = (1,3) $$ so I don't understand where "$|x + 2| \leq |3 + 2| = 5$ for all $x \in V_\delta(c)$" comes from? Any help is much appreciated.

Answer 1

I like using the triangle inequality here instead: $$ |x + 2| = |(x - 2) + (4)| \leq |x - 2| + |4| < 1 + |4| = 5 $$

Answer 2

if $x \in V_{\delta}(c)$, we have $|x-2| < 1$. Since $||x|-|2|| \leq |x-2|$ we deduce: $|x|-2 \leq ||x|-|2|| \leq |x-2| < 1$. Then $|x| < 1+2=3$.

Now : $$|x+2| \leq |x| + |2| < 3+2=5$$