Epsilon delta concept problem

Question

$|a_{n}-a|<\min\left\{1,\frac{\epsilon}{|a|+|b|+1}\right\}$

here what does $\min\left\{1,\frac{\epsilon}{|a|+|b|+1}\right\}$ mean? Similarly, what does $\max\left\{N_{1},N_{2}\right\} $, which i read in a proof mean?

Answer 1

Means that $\delta \leq 1$ AND $\delta \leq \frac{\epsilon}{|a| + |b| + 1}$, so you can use both inequalities when trying to get $|f(x) - L| < \epsilon$. Analogous with $\max$.

Answer 2

An example can clarify better this technic. Suppose that we want to prove that $$\lim_{x\to 0}x^2=0.$$ Let $\varepsilon>0$ arbitrary. We have to find $\delta>0$ that works on the formal definition. Then since we going to use $x$ arbitrary small (near to zero) we can suppose (for simplicity) that $0<|x|<1$. But, also we can suppose that $0<|x|<\varepsilon$, by the same reason. This can be simplify by saying take $\delta=\min\{1,\varepsilon\}$. To conclude the proof note that if $0<|x|<\delta$ then $|x^2|=|x|\cdot|x|<1\cdot |x|<1\cdot \varepsilon=\varepsilon$. And the proof is complete.