Limit-related inequalities with absolute values

Question

Recently I decided to learn calculus on my own and I stumbled across something which I cannot figure why is correct.

Let $f$ be some function for which you know only that if $0<|x-3|<1$, then $|f(x)-5|<0.1$

It is necessary true then that

If $|x-2.5|<0.3$, then $|f(x)-5|<0.1$

I can't for the life of me figure out why that would work.

Answer 1

If $|x-2.5|<0.3$ then $2.2<x<2.8$. Therefore, $2<x<4$, or $|x-3|<1$.

Answer 2

Hint: $|x-2.5|<0.3$ is equivalent to $2.2<x<2.8$

Answer 3

\begin{align} |x - 2.5| < 0.3 & \implies -0.3 < x - 2.5 < 0.3 \\ & \implies -0.8 < x - 3 < -0.2 \\ & \implies |x - 3| < 0.8 \\ & \implies |x - 3| < 1. \end{align}