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

To prove the limit, prove that given $\epsilon\gt0$ there exists a $\delta\gt0$ such that for all $x$ $$0\lt|x-(-3)|\lt\delta\Rightarrow|f(x)-(-6)|\lt\epsilon$$
$$\lim_{x\to-3}\frac{x^2-9}{x+3}=-6$$ Our job is to prove that given $\epsilon\gt0$ there exists a $\delta\gt0$ such that for all $x$ $$0\lt|x-(-3)|\lt\delta\Rightarrow|f(x)-(-6)|\lt\epsilon$$ 1) Solve the inequality $|f(x)-(-6)|\lt\epsilon$ to find an open interval containing $x=-3$ on which the inequality holds for all $x\ne-3$ $$|f(x)-(-6)|\lt\epsilon$$ $$|f(x)+6|\lt\epsilon$$ $$|\frac{x^2-9}{x+3}+6|\lt\epsilon$$ $$|\frac{x^2-9+6x+18}{x+3}|\lt\epsilon$$ $$|\frac{x^2+6x+9}{x+3}|\lt\epsilon$$ $$|\frac{(x+3)^2}{x+3}|\lt\epsilon$$ $$|x+3|\lt\epsilon$$ $$-\epsilon\lt x+3\lt\epsilon$$ $$-\epsilon-3\lt x\lt\epsilon-3$$ So the open interval on which the inequality holds is $(-\epsilon-3,\epsilon-3)$. 2) Find a value of $\delta\gt0$ that places the centered interval $(-3-\delta,-3+\delta)$ inside the interval $(-\epsilon-3,\epsilon-3)$ Take $\delta$ to be the distance from $-3$ to the nearer endpoint of $(-\epsilon-3,\epsilon-3)$. In fact, the distance from $-3$ to both endpoints is the same, which is $\epsilon-3-(-3)=\epsilon$ So if $\delta=\epsilon$ or any smaller positive value, then the constraint $0\lt|x-(-3)|\lt\delta$ will automatically place $x$ between $-\epsilon-3$ and $\epsilon-3$ so that $|f(x)-(-6)|\lt\epsilon$. That means for all $x$, $$0\lt|x-(-3)|\lt\delta\Rightarrow|f(x)-(-6)|\lt\epsilon$$ This completes our proof.