Theorem: If $\lim_{x\to a}\ f(x)=L$ then $\lim_{x\to a}\ cf(x)=cL$.

Suppose $\lim_{x\to a}\ f(x)=L$, and consider any $\epsilon >0$.

To show that $\lim_{x\to a}\ cf(x)=cL$, we must find a $\delta>0$ such that if $0<|x-a|<\delta$ then $|cf(x)-cL|<\epsilon$.

If $c\ne 0$, then $\frac{\epsilon}{|c|}>0$, and since $\lim_{x\to a}\ f(x)=L$, there exists a $\delta>0$ such that if $0< |x-a|< \delta$ then $|f(x)-L|< \frac{\epsilon}{|c|}$.

With this choice of $\delta$, $0<|x-a|<\delta$ 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 $c=0$, we need to prove that $\lim_{x\to a}\ 0=0$, 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:
$\lim_{x\to a}\ (f(x)\ -\ g(x))=\lim_{x\to a}\ f(x)\ -\ \lim_{x\to a}\ g(x)$ ?