Answer
$\lim\limits_{x \to 2} (x^2-4x+5) = 1$
Work Step by Step
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, $\lim\limits_{x \to 2} (x^2-4x+5) = 1$
