Chapter 2 - Section 2.3 - The Precise Definition of a Limit - Exercises - Page 76: 42

To prove the limit, prove that given $\epsilon\gt0$ there exists a $\delta\gt0$ such that for all $x$ $$0\lt|x-(-2)|\lt\delta\Rightarrow|f(x)-4|\lt\epsilon$$

$\lim_{x\to-2}f(x)=4$ if $f(x)=x^2$ for $x\ne-2$ and $f(x)=1$ for $x=-2$ Our job is to prove that given $\epsilon\gt0$ there exists a $\delta\gt0$ such that for all $x$ $$0\lt|x-(-2)|\lt\delta\Rightarrow|f(x)-4|\lt\epsilon$$ 1) Solve the inequality $|f(x)-4|\lt\epsilon$ to find an open interval containing $x=-2$ on which the inequality holds for all $x\ne-2$ (which also means we would consider only $f(x)=x^2$ here) $$|f(x)-4|\lt\epsilon$$ $$|x^2-4|\lt\epsilon$$ $$-\epsilon\lt x^2-4\lt\epsilon$$ $$4-\epsilon\lt x^2\lt4+\epsilon$$ $$\sqrt{4-\epsilon}\lt|x|\lt\sqrt{4+\epsilon}$$ (here we need to assume $\epsilon\lt4$) Also, since we are examining the interval around $x=-2$, we assume that $x\lt0$, meaning $|x|=-x$ $$\sqrt{4-\epsilon}\lt -x\lt\sqrt{4+\epsilon}$$ $$-\sqrt{4-\epsilon}\gt x\gt-\sqrt{4+\epsilon}$$ So the open interval on which the inequality holds is $(-\sqrt{4+\epsilon},- \sqrt{4-\epsilon})$. 2) Find a value of $\delta\gt0$ that places the centered interval $(-2-\delta,-2+\delta)$ inside the interval $(-\sqrt{4+\epsilon},- \sqrt{4-\epsilon})$ Take $\delta$ to be the distance from $-2$ to the nearer endpoint of $(-\sqrt{4+\epsilon},- \sqrt{4-\epsilon})$. In other words, $\delta=\min(-2-(-\sqrt{4+\epsilon}),-\sqrt{4-\epsilon}-(-2))=\min(-2+\sqrt{4+\epsilon},-\sqrt{4-\epsilon}+2)$ So if $\delta$ has this or any smaller positive value, then the constraint $0\lt|x-(-2)|\lt\delta$ will automatically place $x$ between $-\sqrt{4+\epsilon}$ and $-\sqrt{4-\epsilon}$ so that $|f(x)-4|\lt\epsilon$. That means for all $x$, $$0\lt|x-(-2)|\lt\delta\Rightarrow|f(x)-4|\lt\epsilon$$ This completes our proof for $0\lt\epsilon\lt4$. - For $\epsilon\ge4$, we take $\delta$ to be the distance from $-2$ to the nearer endpoint of $(-\sqrt{4+\epsilon},0)$. In other words, $\delta=\min(-2+\sqrt{4+\epsilon},2)$ So again, if $\delta$ has this or any smaller positive value, then the constraint $0\lt|x-(-2)|\lt\delta$ will automatically place $x$ between $-\sqrt{4+\epsilon}$ and $0$ so that $|f(x)-4|\lt\epsilon$. That means for all $x$, $$0\lt|x-(-2)|\lt\delta\Rightarrow|f(x)-4|\lt\epsilon$$ This completes our proof for $\epsilon\ge4$.