Prove $$\lim_{x \to a} cf(x) = c\lim_{x \to a} f(x)$$
Let $\lim_{x \to a} f(x) = L$, where
$$\forall \epsilon > 0, \exists \delta_1 > 0 : (0 < |x-a| < \delta_1 \implies |f(x) - L| < \frac{\epsilon}{|c| + 1})$$
We wish to show that
$$\forall \epsilon > 0, \exists \delta > 0 : (0 < |x-a| < \delta \implies |cf(x) - cL| < \epsilon)$$
If we set $\delta = \delta_1$ then
$$|cf(x) - cL| = |c||f(x) - L| < (|c| + 1)|f(x) - L| < (|c| + 1) \cdot \frac{\epsilon}{|c| + 1} = \epsilon$$
Is this proof correct? Should it work for any real $c$? I had to modify this a little to get it so the denominators wouldn't be $c=0$.
Seems okay after the edit. I propose a structure to an alternate proof:
Let $c \in \mathbb{R}, \ f(x)$ and $g(x)$ be two functions for which the limit exists. After proving the Algebraic Limit Theorem on Multiplication for Function Limits, $$\lim\limits_{x\rightarrow a} \left(f\cdot g\right)(x) = \lim\limits_{x\rightarrow a} g(x) \cdot \lim\limits_{x\rightarrow a} f(x)$$ one can define a constant function $g(x) = c, \forall x \in \mathbb{R}$.
Trivially then $\lim\limits_{x\rightarrow a} g(x) = c$ and hence the result follows. The result is a corollary of an even more general statement. Needless to say, a direct proof is also correct!