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

To prove the limit, prove that given $\epsilon\gt0$ there exists a $\delta\gt0$ such that for all $x$ $$0\lt|x-0|\lt\delta\Rightarrow|f(x)-2|\lt\epsilon$$
$$\lim_{x\to0}\sqrt{4-x}=2$$ Our job is to prove that given $\epsilon\gt0$ there exists a $\delta\gt0$ such that for all $x$ $$0\lt|x-0|\lt\delta\Rightarrow|f(x)-2|\lt\epsilon$$ $$0\lt|x|\lt\delta\Rightarrow|f(x)-2|\lt\epsilon$$ 1) Solve the inequality $|f(x)-2|\lt\epsilon$ to find an open interval containing $x=0$ on which the inequality holds for all $x\ne0$ $$|f(x)-2|\lt\epsilon$$ $$|\sqrt{4-x}-2|\lt\epsilon$$ $$-\epsilon\lt\sqrt{4-x}-2\lt\epsilon$$ $$2-\epsilon\lt\sqrt{4-x}\lt2+\epsilon$$ Square all sides: $$(2-\epsilon)^2\lt 4-x\lt(2+\epsilon)^2$$ $$4-4\epsilon+\epsilon^2\lt 4-x\lt4+4\epsilon+\epsilon^2$$ $$-4\epsilon+\epsilon^2\lt -x\lt 4\epsilon+\epsilon^2$$ $$4\epsilon-\epsilon^2\gt x\gt-4\epsilon-\epsilon^2$$ So the open interval on which the inequality holds is $(-4\epsilon-\epsilon^2, 4\epsilon-\epsilon^2)$. 2) Find a value of $\delta\gt0$ that places the centered interval $(-\delta,\delta)$ inside the interval $(-4\epsilon-\epsilon^2, 4\epsilon-\epsilon^2)$ Take $\delta$ to be the distance from $0$ to the nearer endpoint of $(-4\epsilon-\epsilon^2, 4\epsilon-\epsilon^2)$. In other words, $\delta=\min(0-(-4\epsilon-\epsilon^2),(4\epsilon-\epsilon^2)-0)=\min(4\epsilon+\epsilon^2,4\epsilon-\epsilon^2)$ So if $\delta$ has this or any smaller positive value, then the constraint $0\lt|x|\lt\delta$ will automatically place $x$ between $-4\epsilon-\epsilon^2$ and $4\epsilon-\epsilon^2$ so that $|f(x)-2|\lt\epsilon$. That means for all $x$, $$0\lt|x|\lt\delta\Rightarrow|f(x)-2|\lt\epsilon$$ This completes our proof.