# How do you prove that the limit of (x^2 - 4x + 5) = 1 as x approaches 2 using the epsilon delta proof?

May 10, 2017

See below.

#### Explanation:

Problem:
${\lim}_{x \to 2} \left({x}^{2} - 4 x + 5\right) = 1$

Work:
$| \left({x}^{2} - 4 x + 5\right) - 1 | < \epsilon$
$| {x}^{2} - 4 x + 4 | < \epsilon$
$| \left(x - 2\right) {|}^{2} < \epsilon$
$| x - 2 | < \sqrt{\epsilon}$

Proof:

$\forall$ $\epsilon > 0$, $\exists$ $\delta > 0$ such that
if $0 < | x - 2 | < \delta$, then $| \left({x}^{2} - 4 x + 5\right) - 1 | < \epsilon$
Given $0 < | x - 2 | < \delta$, let $\delta = \sqrt{\epsilon}$
$| x - 2 | < \sqrt{\epsilon}$
$| x - 2 {|}^{2} < \epsilon$
$| {\left(x - 2\right)}^{2} | < \epsilon$
$| {x}^{2} - 4 x + 4 | < \epsilon$
$| \left({x}^{2} - 4 x + 5\right) - 1 | < \epsilon$
$\therefore {\lim}_{x \to 2} \left({x}^{2} - 4 x + 5\right) = 1$