Scalar Multiplication of Limits $\epsilon$ - $\delta $ Proof I am having troubling understanding the $\epsilon$ - $\delta $ proof of the scalar multiplication property of limits, which basically states:
$$\lim_{x\to a}[f(x) \cdot c]=c\cdot L$$
The way I understand it (which doesn't feel to be a good understanding) our choice of $\delta$ is $\frac{\epsilon}{|c|}$, and then do we substitute this value of $\delta$ into the antecedent or the consequent?
Second we are basically trying to satisfy the definition, in this case,  $$|x-a|<\delta \Longrightarrow |c \cdot f(x)-c \cdot L|< \epsilon$$
right? So should I start from this definition(below) and try to work create the above to the above definition(so substitute in the consequent)?
$$|x-a|<\delta \Longrightarrow |  f(x)-  L|< \frac{\epsilon}{|c|}$$
Also, why does this proof use $\delta=\delta_1$??

The proof in the picture is from the following link: http://tutorial.math.lamar.edu/Classes/CalcI/LimitProofs.aspx
 A: You are given that $\lim_{x\rightarrow a}f(x)=L$. That means given any "positive value" there exists "another positive value"(depends on the "positive value") such that 
if $0<|x-a|<$"another positive value" then $|f(x)-L|<$"positive value". This is the fact we have.
Now, you need to show that given any $\epsilon>0$ there exists $\delta>0$, such that $0<|x-a|<\delta$ then $|cf(x)-cL|<\epsilon$. 
So, first take any $\epsilon>0$. Then $\epsilon/|c|$ is also positive. Then by the fact we have, there exists a $\delta>0$ such that if $0<|x-a|<\delta$ then $|f(x)-L|<\epsilon/|c|$. That means $|cf(x)-cL|<\epsilon$. So, we are done.
A: We only choose one $\delta$ throughout this question.
Logically, we first set the error to be $\dfrac\epsilon{|c|}$, then, from $$\lim_{x\to a} f(x)= L,$$ we obtain $\delta_1>0$ such that
$$\color{\red}{0<}|x-a|<\delta_1 \implies |  f(x)-  L|< \frac{\epsilon}{|c|}.$$
Then we set $\delta = \delta_1$, and move $|c|$ to the LHS.
$$\color{\red}{0<}|x-a|<\delta \implies |cf(x)- cL|< \epsilon.$$
Hence, $$\lim_{x\to a}[f(x) \cdot c]=c\cdot L.$$
