## Calculus (3rd Edition)

$$\lim_{x\to 2}x^{-2}=\frac{1}{4}$$
Let $\epsilon\gt 0$ be given. First, we bound $x^{-2}-\frac{1}{4}$ : $$\left|x^{-2}-\frac{1}{4}\right|=\left|\frac{4-x^{2}}{4 x^{2}}\right|=|2-x|\left|\frac{2+x}{4 x^{2}}\right|$$ Let $\delta=\min \left(1, \frac{4}{5} \epsilon\right),$ and suppose $|x-2|\lt \delta .$ Since $\delta\lt 1,|x-2|\lt 1,$ so $1\lt x\lt 3 .$ This means that $4 x^{2}\gt 4$ and $|2+x|\lt 5,$ so that $\frac{2+x}{4 x^{2}}\lt \frac{5}{4} .$ We get: $$\left|x^{-2}-\frac{1}{4}\right|=|2-x|\left|\frac{2+x}{4 x^{2}}\right|\lt \frac{5}{4}|x-2|\lt \frac{5}{4} \cdot \frac{4}{5} \epsilon=\epsilon$$ Hence $$\lim_{x\to 2}x^{-2}=\frac{1}{4}$$