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

$$f(x)=x^2-7\hspace{1cm}c=1$$ For $f$ to be continuous at $c=1$, $\lim_{x\to 1}f(x)=f(1)=1^2-7=-6$ In other words, we need to prove that $\lim_{x\to 1}f(x)=-6$. According to the formal definition of limit, for any values of $\epsilon\gt0$, there must exist a corresponding value of $\delta\gt0$ such that for all $x$, $$0\lt|x-1|\lt\delta\Rightarrow|f(x)-(-6)|\lt\epsilon$$ $$0\lt|x-1|\lt\delta\Rightarrow|f(x)+6|\lt\epsilon$$ Examining the inequality: $$|f(x)+6|\lt\epsilon$$ $$|x^2-7+6|\lt\epsilon$$ $$|x^2-1|\lt\epsilon$$ $$-\epsilon\lt x^2-1\lt\epsilon$$ $$1-\epsilon\lt x^2\lt\epsilon+1$$ As we are considering values of $x$ around $c=1$, we assume that $x\gt0$. Therefore, $$\sqrt{1-\epsilon}\lt x\lt\sqrt{\epsilon+1}$$ So if $x\in(\sqrt{1-\epsilon},\sqrt{\epsilon+1})$ then $|f(x)+6|\lt\epsilon$ Take $\delta$ to be the distance from $c=1$ to the nearer endpoint of $(\sqrt{1-\epsilon},\sqrt{\epsilon+1})$, or $\delta=\min(1-\sqrt{1-\epsilon},\sqrt{\epsilon+1}-1)$. If $\delta$ has this value or any smaller values, then because $0\lt|x-1|\lt\delta$, $x$ will always lie in the interval $(\sqrt{1-\epsilon},\sqrt{1+\epsilon})$ so that $|f(x)+6|\lt\epsilon$. Therefore, $\lim_{x\to1}f(x)=f(1)=-6$. $f$ is continuous at $x=1$.