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

We can say $\lim_{x\to c}f(x)=L$ if for every number $\epsilon\gt0$, there exists a corresponding number $\delta\gt0$ such that for all $x$, $$0\lt |x-c|\lt\delta\Rightarrow |f(x)-L|\lt\epsilon$$
We can say $\lim_{x\to c}f(x)=L$ if for every number $\epsilon\gt0$, there exists a corresponding number $\delta\gt0$ such that for all $x$, $$0\lt |x-c|\lt\delta\Rightarrow |f(x)-L|\lt\epsilon$$ Take this function: $f(x)=x-1$ We know that $\lim_{x\to1}f(x)=1-1=0$ According to the precise definition of limit, it must be the case that for every number $\epsilon\gt0$, there exists a corresponding number $\delta\gt0$ such that for all $x$, $$0\lt|x-1|\lt\delta\Rightarrow|f(x)|\lt\epsilon$$ To find $\delta$, we consider the range of $x$ that the inequality $|f(x)|\lt\epsilon$ is satisfied. $$|x-1|\lt\epsilon$$ $$-\epsilon\lt x-1\lt\epsilon$$ So if we take $\delta=\epsilon$ in this case, then $|x-1|\lt\epsilon$, which means $|f(x)|\lt\epsilon$. We have found $\delta$ that satisfies the inequality $|f(x)|\lt\epsilon.$