#### Answer

Prove that for every positive real number $B$, there exists a corresponding $\delta\gt0$ such that for all $x$ $$2\lt x\lt 2+\delta\Rightarrow \frac{1}{x-2}\gt B$$

#### Work Step by Step

$$\lim_{x\to2^+}\frac{1}{x-2}=\infty$$
According to the formal definition, we need to prove that for every positive real number $B$, there exists a corresponding $\delta\gt0$ such that for all $x$ $$2\lt x\lt 2+\delta\Rightarrow \frac{1}{x-2}\gt B$$
- Examine the inequality: $$\frac{1}{x-2}\gt B$$ $$x-2\lt\frac{1}{B}$$ $$x\lt2+\frac{1}{B}$$
- So if we set $\delta=\frac{1}{B}$, then $2\lt x\lt2+\frac{1}{B}$, we would for all $x$ have $\frac{1}{x-2}\gt B$, satisfying the requirements.
The proof has been completed.