## Thomas' Calculus 13th Edition

Step 1. To prove $\lim_{x\to 1^-}\frac{1}{1-x^2}=\infty$ , for every positive real number $B$, we need to find a corresponding number $\delta\gt0$ such that for all x, $-\delta\lt x-1 \lt 0$, we get $\frac{1}{1-x^2}\gt B$ Step 2. The last inequality gives $1-x^2 \lt \frac{1}{B}$ or $x^2\gt1-\frac{1}{B}$ which gives $|x|\gt\sqrt {1-1/B}$, thus we can choose $\delta=1-\sqrt {1-1/B}$ so that when we go back in the steps, we see that for all x, $-\delta\lt x-1 \lt 0$, we get $\frac{1}{1-x^2}\gt B$ which proves the limit statement.