Prove that $\lim_{x\to a}\bigr(C f(x)\bigl) = C \lim_{x\to a} f(x)$

Question

I would like to prove the following:

Prove that $\lim_{x\to a}\bigr(C f(x)\bigl) = C \lim_{x\to a} f(x)$, where $C$ is constant.

First I checked some proofs on internet. One of the proofs that I came across is this (Property $1$). (Click on the image to expand)

However, considering that this property is quite basic, I wonder whether there is a shorter (and less cryptic) way to prove that.

My feeble attempt was:

According to the Wikipedia:

A limit is the value that a function (or sequence) "approaches" as the input (or index) "approaches" some value, or in other words:

$$\lim_{x\to a} f(x) = L$$ where $L$ is some constant.

Since $C$ is constant and unrelated to value of $x$, when $x$ approaches $a$, $C$ remains unchanged and $f(x)$ approaches $L$, in other words:

$\lim_{x\to a}\bigr(C f(x)\bigl) = CL$, which can be rewritten as $C \lim_{x\to a} f(x)$

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Very likely that my attempt is incorrect, hence I would like to ask you, is there a better (and shorter) alternative to show that $\lim_{x\to a}\bigr(C f(x)\bigl) = C \lim_{x\to a} f(x)$

Answer 1

As $|Cf(x)-CL|=|C||f(x)-L|$, the simplest and possibly only way to complete the proof with the $\epsilon$ - $\delta$ terminology is to let $|f(x)-L|<\frac{\epsilon}{|C|}$ for every $\epsilon>0$. I don't see a way that is shorter than the pasted image unless you have already proven $3$ and $7$.

Answer 2

What you need to show is that if for every $\epsilon>0$ there is $\delta>0$ such that $|f(x)-L|<\epsilon$ whenever $|x-a|<\delta$, then for every $\epsilon>0$ there is $\delta>0$ such that $|Cf(x)-CL|<\epsilon$ whenever $|x-a|<\delta$.