## Calculus: Early Transcendentals 8th Edition

$\lim\limits_{x \to 2} (x^2-4x+5) = 1$
Let $\epsilon \gt 0$ be given. Let $\delta = \sqrt{\epsilon}$ Suppose that $\vert x-2 \vert \lt \delta$ Then: $\vert (x^2-4x+5) - 1\vert = \vert x^2-4x+4\vert = \vert (x-2)^2\vert\lt \delta^2 = \epsilon$ Therefore, $\lim\limits_{x \to 2} (x^2-4x+5) = 1$