Proof of Various Limit Properties — Question 6

Question 6 - Solution

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

Simplify the expression: $$|(f(x)-c)-(L-c)| = |f(x)-L|.$$

Since $\lim_{x\to a} f(x)=L$, by the definition of the limit there exists $\delta>0$ such that whenever $$0<|x-a|<\delta,$$ we have $$|f(x)-L|<\varepsilon.$$

Thus, for this same choice of $\delta$, $$|(f(x)-c)-(L-c)|<\varepsilon.$$

Therefore, $$\lim_{x\to a} \bigl(f(x)-c\bigr)=L-c.$$