Proof of Various Limit Properties — Question 9

Question 9 - Solution

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

Begin by rewriting the difference: $$f(x)-g(x) = \bigl(f(x)-L\bigr)-\bigl(g(x)-L\bigr).$$

Take absolute values and apply the triangle inequality: $$|f(x)-g(x)| \le |f(x)-L|+|g(x)-L|.$$

Since $\lim_{x\to a} f(x)=L$, there exists $\delta_1>0$ such that $$|f(x)-L|<\frac{\varepsilon}{2} \quad\text{whenever }0<|x-a|<\delta_1.$$

Similarly, since $\lim_{x\to a} g(x)=L$, there exists $\delta_2>0$ such that $$|g(x)-L|<\frac{\varepsilon}{2} \quad\text{whenever }0<|x-a|<\delta_2.$$

Let $$\delta=\min(\delta_1,\delta_2).$$

Then for all $x$ satisfying $0<|x-a|<\delta$, $$|f(x)-g(x)| \le |f(x)-L|+|g(x)-L| < \frac{\varepsilon}{2}+\frac{\varepsilon}{2} =\varepsilon.$$

Therefore, $$\lim_{x\to a}\bigl(f(x)-g(x)\bigr)=0.$$