How is this proof for the scalar product rule of limits valid?

Question

If we let $K=\lim\limits_{x \to a} f(x),$ and let $c$ be a constant,

Then in order to show that $\lim\limits_{x \to a} cf(x) = cK$, we must show that there is an $\epsilon$ for every $\delta$ such that

$\lvert cf(x)-cK \rvert < \epsilon$ whenever $\lvert x-a \rvert < \delta$.

The proof claims to prove this by staing:

$$\lvert cf(x)-cK \rvert = \lvert c \rvert \lvert f(x)-K \rvert < \lvert c \rvert \frac{\epsilon}{\lvert c \rvert} = \epsilon.$$

However, I don't see how this proves anything other than basic manipulation of terms, and I don't see how it relates $\epsilon$ to $\delta$ in any way.

Answer 1

For every value of $\varepsilon>0,$ there exists $\delta>0$ such that if $0<|x-a|<\delta$ then $|f(x) - K|<\varepsilon.$

What does “every” mean?

For every positive number $\varepsilon$, blah blah blah etc. is true of $\varepsilon.$

Is $\varepsilon/|c|$ a positive number?

If so then blah blah blah etc. is true of $\varepsilon/|c|.$

Thus there exists $\delta>0$ such that if $0<|x-a|<\delta$ then $|f(x)-K| < \dfrac \varepsilon {|c|}.$

(This generally requires a smaller value of $\delta$ than what would be needed if we'd just wanted $|f(x)-K|<\varepsilon.$ But we know that such a value of $\delta$ exists because of the word "every".)

And then go through the rest of the argument.

Answer 2

The step missing is saying "So take $\delta>0$ such that $|f(x)-K|<\epsilon/|c|$ whenever $|x-a|<\delta$. Thus, that $\epsilon$ and $\delta$ will work to show that $|cf(x)-cK|<\epsilon$ whenever $|x-a|<\delta$."

Most of the time, this step is left out of proof involving limits since it has the same flavor and is largely the same every time.

One other thing worth mentioning is that this proof doesn't work when $c=0$, but that exceptional case is rather trivial since it just leads to the claim that $\lim_{x \to a}{0}=0$.