#### Answer

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

#### Work Step by Step

$$\lim_{x\to1^-}\frac{1}{1-x^2}=\infty$$
According to the formal definition deduced from Exercise 93, we need to prove that for every positive real number $B$, there exists a corresponding $\delta\gt0$ such that for all $x$ $$1-\delta\lt x\lt 1\Rightarrow \frac{1}{1-x^2}\gt B$$
- Examine the inequality: $$\frac{1}{1-x^2}\gt B$$ $$1-x^2\lt\frac{1}{B}$$ $$x^2\gt1-\frac{1}{B}$$
- So if we set $\delta=\frac{1}{2B}$, then $$1-\frac{1}{2B}\lt x\lt 1$$ $$1-2\times\frac{1}{2B}+\frac{1}{4B^2}\lt x^2\lt1$$ $$1-\frac{1}{B}+\frac{1}{4B^2} \lt x^2\lt1$$
Since $\frac{1}{4B^2}\gt0$, $x^2\gt1-\frac{1}{B}+\frac{1}{4B^2}\gt1-\frac{1}{B}$, meaning that we would for all $x$ have $\frac{1}{1-x^2}\gt B$, satisfying the requirements.
The proof has been completed.