# How do you prove that the limit of (x^2 - 1) = 3 as x approaches -2 using the epsilon delta proof?

Jun 4, 2017

See below.

#### Explanation:

Prove ${\lim}_{x \to - 2} \left({x}^{2} - 1\right) = 3$

Work (not part of proof):
$0 < | x + 2 | < \delta$; $| \left({x}^{2} - 1\right) - 3 | < \epsilon$

We need to manipulate the $| \left({x}^{2} - 1\right) - 3 | < \epsilon$ to show that $| x + 2 | < \text{something}$ to set delta equal to that term:

$| \left({x}^{2} - 1\right) - 3 | < \epsilon$

$| {x}^{2} - 4 | < \epsilon$

$| \left(x + 2\right) \left(x - 2\right) | < \epsilon$

$| x + 2 | < \frac{\epsilon}{x - 2}$

Since we cannot have a $x$ term with epsilon, we let $\delta = 1$ and solve for the value $x + 2$ would be:
$0 < | x + 2 | < 1$

$- 1 < x + 2 < 1$

$- 1 - 4 < x + 2 - 4 < 1 - 4$

$- 5 < x - 2 < - 3$

Here, we choose the larger value since if we chose the smaller value, the -5 would not be included, so:
$| x - 2 | < 5$

Therefore,
$| x + 2 | < \frac{\epsilon}{5}$

Proof:
$\forall$ $\epsilon > 0$, $\exists$ $\delta > 0$ such that:
if $0 < | x + 2 | < \delta$, then $| \left({x}^{2} - 1\right) - 3 | < \epsilon$.
Given $0 < | x + 2 | < \delta$, let $\delta = \min \left(1 , \frac{\epsilon}{5}\right)$:
$0 < | x + 2 | < \frac{\epsilon}{5}$

$0 < 5 | x + 2 | < \epsilon$

$0 < | x - 2 | | x + 2 | < \epsilon$

$0 < | {x}^{2} - 4 | < \epsilon$

$0 < | \left({x}^{2} - 1\right) - 3 | < \epsilon$

$\therefore$ ${\lim}_{x \to - 2} \left({x}^{2} - 1\right) = 3$