Proof of Various Limit Properties — Question 3

Question 3 - Solution

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

Recall the inequality $$\bigl||u|-|v|\bigr| \le |u-v|$$ for all real numbers $u$ and $v$.

Apply this inequality with $u=f(x)$ and $v=L$: $$\bigl||f(x)|-|L|\bigr| \le |f(x)-L|.$$

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

Combining the inequalities gives $$\bigl||f(x)|-|L|\bigr| \le |f(x)-L| < \varepsilon.$$

Thus, whenever $0<|x-a|<\delta$, $$\bigl||f(x)|-|L|\bigr|<\varepsilon.$$

Therefore, $$\lim_{x\to a} |f(x)| = |L|.$$