$2a-1< x+a < 2a+1 \implies |x+a| < 2|a|+1$

Question

In tried to do a rigorous delta-epsilon argument for $\lim_{x \to a} x^2 = a^2$. I found this post, which I uses the same way as I did. But I don't fully understand on of the steps:

If $|x-a| < 1$ then $ -1 < x-a <1 $ hence $ a-1 < x < a+1$ therefore $2a-1< x+a <2a+1$. No how do we arrive at $|x+a| < 2|a|+1$?

Intuitively and geometrically on the real line I think I know this to be true. But is this enough for a formal proof, or is there some algebraic identity, etc, that I don't see, that justifies this step?

I know the identity $\left| a \right| - \left| b \right| \leq \left| a -b \right|$, which leads to the same result and is used in a answer to the aforementioned post. But is there a direct way to deduce this from $2a-1< x+a <2a+1$?

I thought about $\left| x \right| \lt a \iff -a \lt x \lt a$. But to use this for $|x+a| < 2|a|+1$ I think we would need $-(2a-1) \lt x \lt 2a+1$ rather than $2a-1 < x+a <2a+1$.

Am I getting something wrong? Sorry if this is somewhat confused. I am just starting to learn this kind of math. Any help appreciated.

Answer 1

If $|x-a| < 1$ then $ -1 < x-a <1 $ hence $ a-1 < x < a+1$ therefore $2a-1< x+a <2a+1$. No how do we arrive at $|x+a| < 2|a|+1$?

Note $|x+a| < 2|a|+1$ means $-2|a|-1<x+a<2|a|+1$.

Since you already have $ 2a-1< x+a <2a+1$, now it is easy to verify the following

$$-2|a|-1\le 2a-1<x+a<2a+1\le2|a|+1$$

Therefore, you arrive at

$$|x+a| < 2|a|+1$$

Answer 2

$|x+a| \leq |x|+|a|$ ; Triangle inequality

then use the reverse triangle inequality as you mentioned,

$|x|-|a| \leq |x-a| < 1$ which implies $|x| < 1+|a|$

Therefore,

$|x+a| \leq |x|+|a| < 2|a|+1$

Answer 3

$$2a-1< x+a < 2a+1 \implies |x+a| < \max\{|2a-1|,|2a+1|\}$$

By triangle inequality, $|2a-1|\le|2a|+|-1|=2|a|+1$ and $|2a+1|\le|2a|+|1|=2|a|+1$ so $|x+a|<2|a|+1$