## University Calculus: Early Transcendentals (3rd Edition)

To prove the limit, prove that given $\epsilon\gt0$ there exists a $\delta\gt0$ such that for all $x$ $$0\lt|x-\sqrt3|\lt\delta\Rightarrow|f(x)-\frac{1}{3}|\lt\epsilon$$
$$\lim_{x\to\sqrt3}\frac{1}{x^2}=\frac{1}{3}$$ Our job is to prove that given $\epsilon\gt0$ there exists a $\delta\gt0$ such that for all $x$ $$0\lt|x-\sqrt3|\lt\delta\Rightarrow|f(x)-\frac{1}{3}|\lt\epsilon$$ 1) Solve the inequality $|f(x)-\frac{1}{3}|\lt\epsilon$ to find an open interval containing $x=\sqrt3$ on which the inequality holds for all $x\ne\sqrt3$ $$|f(x)-\frac{1}{3}|\lt\epsilon$$ $$|\frac{1}{x^2}-\frac{1}{3}|\lt\epsilon$$ $$-\epsilon\lt \frac{1}{x^2}-\frac{1}{3}\lt\epsilon$$ $$\frac{1}{3}-\epsilon\lt \frac{1}{x^2}\lt\frac{1}{3}+\epsilon$$ $$\frac{1-3\epsilon}{3}\lt\frac{1}{x^2}\lt\frac{1+3\epsilon}{3}$$ $$\frac{3}{1-3\epsilon}\gt x^2\gt\frac{3}{1+3\epsilon}$$ Considering the fact that we are examining the interval containing $x=\sqrt3$, we assume that $x\gt0$, so that $\sqrt{x^2}=|x|=x$ Take the square root of all sides: $$\sqrt{\frac{3}{1-3\epsilon}}\gt x\gt\sqrt{\frac{3}{1+3\epsilon}}$$ We need to assume here that $\frac{3}{1-3\epsilon}\gt0\Rightarrow1-3\epsilon\gt0\Rightarrow-3\epsilon\gt-1\Rightarrow\epsilon\lt1/3$ So the open interval on which the inequality holds is $(\sqrt{\frac{3}{1+3\epsilon}},\sqrt{\frac{3}{1-3\epsilon}})$. 2) Find a value of $\delta\gt0$ that places the centered interval $(\sqrt3-\delta,\sqrt3+\delta)$ inside the interval $(\sqrt{\frac{3}{1+3\epsilon}},\sqrt{\frac{3}{1-3\epsilon}})$ Take $\delta$ to be the distance from $\sqrt3$ to the nearer endpoint of $(\sqrt{\frac{3}{1+3\epsilon}},\sqrt{\frac{3}{1-3\epsilon}})$. In other words, $\delta=\min(\sqrt3-\sqrt{\frac{3}{1+3\epsilon}},\sqrt{\frac{3}{1-3\epsilon}}-\sqrt3)$ So if $\delta$ has this or any smaller positive value, then the constraint $0\lt|x-\sqrt3|\lt\delta$ will automatically place $x$ between $\sqrt{\frac{3}{1+3\epsilon}}$ and $\sqrt{\frac{3}{1-3\epsilon}}$ so that $|f(x)-1/3|\lt\epsilon$. That means for all $x$, $$0\lt|x-\sqrt3|\lt\delta\Rightarrow|f(x)-\frac{1}{3}|\lt\epsilon$$ This completes our proof for $0\lt\epsilon\lt1/3$. - For $\epsilon\ge1/3$: Take $\delta$ to be the distance from $\sqrt3$ to the nearer endpoint of $(0,\sqrt{\frac{3}{1+3\epsilon}},)$. In other words, $\delta=\min(\sqrt3,\sqrt3-\sqrt{\frac{3}{1+3\epsilon}})$ So again, if $\delta$ has this or any smaller positive value, then the constraint $0\lt|x-\sqrt3|\lt\delta$ will automatically place $x$ between $0$ and $\sqrt{\frac{3}{1+3\epsilon}}$ so that $|f(x)-1/3|\lt\epsilon$. That means for all $x$, $$0\lt|x-\sqrt3|\lt\delta\Rightarrow|f(x)-\frac{1}{3}|\lt\epsilon$$ This completes our proof for $\epsilon\ge1/3$.