Proof of Various Limit Properties — Question 2

Question 2 - Solution

Let $\varepsilon>0$ be given. We must show that there exists $\delta>0$ such that if $$0<|x-a|<\delta,$$ then $$|cf(x)-cL|<\varepsilon.$$

Start by factoring out the constant: $$|cf(x)-cL| = |c|\cdot|f(x)-L|.$$

Since $\lim_{x\to a} f(x)=L$, there exists $\delta>0$ such that whenever $$0<|x-a|<\delta,$$ we have $$|f(x)-L|<\frac{\varepsilon}{|c|},$$ provided $c\neq 0$.

Substitute this into the inequality: $$|cf(x)-cL| = |c|\cdot|f(x)-L| < |c|\cdot\frac{\varepsilon}{|c|} = \varepsilon.$$

Thus, whenever $0<|x-a|<\delta$, we have $$|cf(x)-cL|<\varepsilon.$$

Therefore, $$\lim_{x\to a} \bigl(cf(x)\bigr)=cL.$$

If $c=0$, then $cf(x)=0$ for all $x$, and the result holds trivially.