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

1) $L=\lim_{x\to2}f(x)=4$ 2) $\delta=0.05$
$$f(x)=\frac{x^2-4}{x-2}\hspace{1cm}c=2\hspace{1cm}\epsilon=0.05$$ 1) Find $L=\lim_{x\to2}f(x)$ $$\lim_{x\to2}f(x)=\lim_{x\to2}\frac{x^2-4}{x-2}=\lim_{x\to2}\frac{(x-2)(x+2)}{x-2}=\lim_{x\to2}(x+2)$$ $$\lim_{x\to2}f(x)=2+2=4$$ Therefore, $L=\lim_{x\to2}f(x)=4$ 2) Find $\delta\gt0$ such that for all $x$, $0\lt |x-2|\lt\delta$ then $|f(x)-4|\lt0.05$ - First, solve the inequality to figure out the open interval on which the inequality holds. $$|f(x)-4|\lt0.05$$ $$|\frac{x^2-4}{x-2}-4|\lt0.05$$ $$|\frac{x^2-4-4x+8}{x-2}|\lt0.05$$ $$|\frac{x^2-4x+4}{x-2}|\lt0.05$$ $$|\frac{(x-2)^2}{x-2}|\lt0.05$$ $$|x-2|\lt0.05$$ $$-0.05\lt x-2\lt0.05$$ $$1.95\lt x\lt2.05$$ The open interval on which the inequality holds is $(1.95,2.05)$. - Find a value of $\delta\gt0$ The distance from $2$ to both endpoints is the same, which is $0.05$. So if we take $\delta=0.05$, then $0\lt|x-2|\lt0.05$, meaning all $x$ now are placed between $1.95$ and $2.05$, hence $|f(x)-4|\lt0.05$ In other words, $$0\lt|x-2|\lt0.05\Rightarrow|f(x)-4|\lt0.05$$