Answer
Prove that for every negative real number $-B$, there exists a corresponding $\delta\gt0$ such that for all $x$ $$2-\delta\lt x\lt 2\Rightarrow \frac{1}{x-2}\lt -B$$
Work Step by Step
$$\lim_{x\to2^-}\frac{1}{x-2}=-\infty$$
According to the formal definition deduced from Exercise 93, we need to prove that for every negative real number $-B$, there exists a corresponding $\delta\gt0$ such that for all $x$ $$2-\delta\lt x\lt 2\Rightarrow \frac{1}{x-2}\lt -B$$
- Examine the inequality: $$\frac{1}{x-2}\lt -B$$ $$x-2\gt-\frac{1}{B}$$ $$x\gt2-\frac{1}{B}$$
- So if we set $\delta=\frac{1}{B}$, then we have $2-\frac{1}{B}\lt x\lt 2$, meaning that we would for all $x$ have $\frac{1}{x-2}\lt -B$, satisfying the requirements.
The proof has been completed.
