Answer
Please see below.
Work Step by Step
We want to prove that $$\lim_{x \to c}[bf(x)]=bL$$ by using $\epsilon - \delta$ definition; that is, we must show that for each $\epsilon >0$, there exists a $\delta >0$ such that $|bf(x)-bL|< \epsilon$ whenever $|x-c|< \delta$.
By the assumption, $\lim_{x \to c} f(x) =L$; that is, for any $\epsilon > 0$, there exists a $\delta >0$ such that$$|x-c|< \delta \quad \Rightarrow \quad |f(x)-L|< \epsilon .$$So by choosing $\epsilon = \frac{\epsilon }{|b|}>0$ we conclude that there exists $\delta _1 >0$ such that$$|x-c|< \delta _1 \quad \Rightarrow \quad |f(x)-L|< \frac{\epsilon }{|b|}.$$Now by noting that $|bf(x)-bL|=|b(f(x)-L)|=|b||f(x)-L|$ we conclude that the statement is true.
