Step in conservation of inequalities and limits

Question

I am a bit confused about the following step in a proof:

for all $\epsilon>0$ we have that $b<c+2 \epsilon$, we can conclude that $b\leq c$.

The proof goes as follows: We wish to prove that if $f(x) \leq g(x)$ and both $\lim_{x\to a} f(x)=b$ and $\lim_{x\to a} g(x)=c$, we have that $b\leq c$.

The author starts out with noticing that since both limits exist we can say that whenever

$|x-a|<\delta_1$, we can say $|f(x)-b|<\epsilon$, thus $b<f(x)+ \epsilon$.

Also notice that whenever

$|x-a|<\delta_2$, we can say $|g(x)-c|<\epsilon$, thus $g(x)-\epsilon < c$ and $g(x)< c+ \epsilon$.

Let's combine these two facts and say that $b<f(x)+\epsilon \leq g(x) + \epsilon < c + 2\epsilon$. I agree with all these steps and have arrived at this myself, I just can't get why the author jumps to the conclusion that $b \leq c$. What am I missing?

Answer 1

This is true for all $\varepsilon >0$. Now, if $\forall \varepsilon >0,\;b-c<2\varepsilon$, then $b-c$ is less than every positive real number, thus $b-c\le 0$, i.e $b\le c$.

Answer 2

If for all $\epsilon > 0$ we have $b < c + 2 \epsilon$, then we can conclude that $b \le c$.

Suppose otherwise that $b > c$. Can you find an $\epsilon$ such that $b \not < c + 2 \epsilon$?