Proof of Various Limit Properties — Question 5

Question 5 - Solution

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

Simplify the expression inside the absolute value: $$|(f(x)+c)-(L+c)| = |f(x)-L|.$$

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

Therefore, for the same $\delta$, $$|(f(x)+c)-(L+c)|<\varepsilon.$$

This proves that $$\lim_{x\to a} \bigl(f(x)+c\bigr)=L+c.$$