Suppose $\lim_{x\to a}\ f(x)=L,ドル and consider any $\epsilon>0$.
To show that $\lim_{x\to a}\ cf(x)=cL,ドル we must find a $\delta>0$ such that if 0ドル<|x-a|<\delta$ then $|cf(x)-cL|<\epsilon$.
If $c\ne 0,ドル then $\frac{\epsilon}{|c|}>0,ドル and since $\lim_{x\to a}\ f(x)=L,ドル there exists a $\delta>0$ such that if 0ドル< |x-a|< \delta$ then $|f(x)-L|< \frac{\epsilon}{|c|}$.
With this choice of $\delta,ドル 0ドル<|x-a|<\delta$ implies $|cf(x)-cL|=|c||f(x)-L|< |c|(\frac{\epsilon}{|c|})=\epsilon$.
This proves that $\lim_{x\to a}\ cf(x)=cL$ when $c\ne 0$.
In the case $c=0,ドル we need to prove that $\lim_{x\to a}\ 0=0,ドル which follows easily from the more general result that $\lim_{x\to a}\ c=c$.
Another way to write this law is: $\lim_{x\to a}\ cf(x)=c\lim_{x\to a}\ f(x)$.
Note that in mathematics, an $=$ sign automatically means that the quantities on both sides of the equality must exist and are equal, so this version of the law assumes (implicitly) that $\lim_{x\to a}\ f(x)$ exists. In the version stated at the top of this page, this assumption is made explicit.
Q: Can you see how this limit law together with the Sum of
Limits$=$Limit of the Sum law implies the Difference of Limits Law:
$\lim_{x\to a}\ (f(x)\ -\ g(x))=\lim_{x\to a}\ f(x)\ -\ \lim_{x\to a}\ g(x)$ ?