Trying to prove $\lim_{x \to a} c f(x) = c \lim_{x \to a} f(x)$ I want to prove that $\lim_{x \to a} c f(x) = c \lim_{x \to a} f(x)$ for nonzero $c$.
Let $\lim_{x \to a} f(x) = L$. Then there exists a $\delta$ such that $|f(x) - L| < \epsilon$ whenever $0 < |x-a| < \delta$.
Since we have $c\lim_{x \to a} f(x) = cL$ this also means $\lim_{x \to a} c f(x) = c L$ from the problem statement.
So there also exists a $\delta_1$ such that $|cf(x) - cL| < \epsilon_1$ whenever $0<|x-a|<\delta_1$. Dividing by $|c|$ we rephrase: There also exists a $\delta_1$ such that $|f(x) - L| < \frac{\epsilon_1}{|c|}$ whenever $0<|x-a|<\delta_1$.
Stuck here. I know we have $|f(x)-L|$ in both pieces now but all this tells me is that $\delta$ implies $|f(x) - L| < \epsilon$ and $\delta_1$ implies $|f(x) - L| < \frac{\epsilon_1}{|c|}$. I don't think this is enough to show they are related or equivalent.
Is it enough to say let $\epsilon_1 = |c| \epsilon$ and let $\delta_1 = \delta$? Are we even allowed to do that?
 A: We are given that $\lim_{x \to a} f(x) = L$ that is


*

*$\exists \delta$ such that $|f(x) - L| < \epsilon$ whenever $0 <
   |x-a| < \delta$


and we have to show that $\lim_{x \to a} c f(x) = c L$.
To prove it we need to find 


*

*$\delta_1$ such that $|cf(x) - cL| < \epsilon_1$ whenever $0 <
   |x-a| < \delta_1$


and since


*

*$|cf(x) - cL| < \epsilon_1 \iff |f(x) - L| < \frac{\epsilon_1}{|c|}$


by the definition of limit it suffices set $\epsilon_3=\frac{\epsilon_1}{|c|}$ to assure that $\exists \delta_3$ such that $|f(x) - L| < \epsilon_3$ whenever $0 <
   |x-a| < \delta_3$.
Then it suffice assume $\delta_1=\delta_3$.
A: Assume $c \neq 0$, since for $c=0$ it is trivial. Start letting $\epsilon > 0$. Since $\lim_{x \to a}f(x) = L$, there is $\delta > 0$ such that $$0 < |x-a|<\delta \implies |f(x)-L| < \color{red}{\epsilon/|c|}.$$The point is that since we already know that $\lim_{x \to a}f(x)=L$, we can put anything positive in that red "placeholder". Now we claim that this $\delta$ works. Assume that $0<|x-a|<\delta$. Then $$|cf(x) - cL| = |c||f(x)-L| < |c| (\epsilon/|c|) = \epsilon.$$This proves that $\lim_{x \to a}cf(x) = cL$.
