Picking a $\delta$ for a convenient $\varepsilon$?

Question

I'm studying some proofs in a book from the library and unlike my own book, it doesn't prove by showing that a certain expression is striclty less than $\varepsilon$ (given an appropiate $\delta$ and $\vert x-a\vert < \delta$, etc). Instead, it shows that it's less than some expression involving $\varepsilon$.

E.g. Given $\varepsilon > 0$, we can pick a $\delta > 0$ so $|f(x) - g(a)| < \text{insert stuff} < k \cdot \varepsilon$ when $|x-a| < \delta$, and thus $f(x) - g(a) \to 0 $ when $x \rightarrow a$.

How does this make sense formally? (Intuitively, I am convinced.)

I don't have much experience with $\varepsilon$-$\delta$ -proofs, so I might be missing something obvious.

Answer 1

To make sense of this formally we do this: You know that when $0<\vert x-a\vert< \delta$, we have $\vert f(x)-g(a)\vert <k \varepsilon$. Now we simply say that $\varepsilon'=\varepsilon \cdot k$.

Now we can say that for every $\varepsilon '>0$ there is a $ \delta<0$, such that ...etc. This is the formal definition of a limit as you know it, which proves the equality of the statements.

Answer 2

You can think of $\epsilon$-$\delta$ proofs as a game: you give me an $\epsilon$ and I'll find you a $\delta$ so that $|f(x)-f(y)| < \epsilon$ whenever $|x-y|<\delta$.

Suppose my proof gives you a $\delta$ so that $|f(x)-f(y)| < k\cdot \epsilon$, then instead of giving me $\epsilon$, you can instead give me $\epsilon/k$. This way, I will give you a $\delta$ so that $|f(x)-f(y)|< \epsilon$. The important thing here is that $k$ cannot depend on $\epsilon$ but must be fixed, so that in our game, you know beforehand what number to really give me (e.g., instead of $\epsilon$, you are giving me $\epsilon/k$) so that $|f(x)-f(y)| < \epsilon$.