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

To prove the limit, prove that given $\epsilon\gt0$ there exists a $\delta\gt0$ such that for all $x$ $$0\lt|x-1|\lt\delta\Rightarrow|f(x)-1|\lt\epsilon$$
$$\lim_{x\to1}\frac{1}{x}=1$$ Our job is to prove that given $\epsilon\gt0$ there exists a $\delta\gt0$ such that for all $x$ $$0\lt|x-1|\lt\delta\Rightarrow|f(x)-1|\lt\epsilon$$ 1) Solve the inequality $|f(x)-1|\lt\epsilon$ to find an open interval containing $x=1$ on which the inequality holds for all $x\ne1$ $$|f(x)-1|\lt\epsilon$$ $$|\frac{1}{x}-1|\lt\epsilon$$ $$-\epsilon\lt \frac{1}{x}-1\lt\epsilon$$ $$1-\epsilon\lt \frac{1}{x}\lt1+\epsilon$$ $$\frac{1}{1-\epsilon}\gt x\gt\frac{1}{1+\epsilon}$$ So the open interval on which the inequality holds is $(\frac{1}{1+\epsilon},\frac{1}{1-\epsilon})$. 2) Find a value of $\delta\gt0$ that places the centered interval $(1-\delta,1+\delta)$ inside the interval $(\frac{1}{1+\epsilon},\frac{1}{1-\epsilon})$ Take $\delta$ to be the distance from $1$ to the nearer endpoint of $(\frac{1}{1+\epsilon},\frac{1}{1-\epsilon})$. In other words, $\delta=\min(1-\frac{1}{1+\epsilon},\frac{1}{1-\epsilon}-1)$ So if $\delta$ has this or any smaller positive value, then the constraint $0\lt|x-1|\lt\delta$ will automatically place $x$ between $1/(1+\epsilon)$ and $1/(1-\epsilon)$ so that $|f(x)-1|\lt\epsilon$. That means for all $x$, $$0\lt|x-1|\lt\delta\Rightarrow|f(x)-1|\lt\epsilon$$ This completes our proof.