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

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