# How do you prove the statement lim as x approaches 4 for  (x^2) = 4 using the epsilon and delta definition?

Oct 15, 2015

Note that the limit as x approaches 4 is 16, and the limit as x approaches 2 is 4.
I will prove the latter case for you.

#### Explanation:

By definition, ${\lim}_{x \to {x}_{0}} f \left(x\right) = L \iff \forall \epsilon > 0 , \exists \delta > 0$ such that $| x - {x}_{0} | < \delta \implies | f \left(x\right) - L | < \epsilon$

Let $\epsilon > 0$ and select $k \in {\mathbb{R}}^{+}$ such that $| x - 2 | < k$

$\therefore - k < x - 2 < k$

$\therefore 4 - k < x + 2 < k + 4$

Hence $| x + 2 | < k + 4$

Now select delta=min{epsilon/(k+4); k}

Clearly $\delta > 0$

Let $| x - 2 | < \delta$

Now : $| f \left(x\right) - 4 | = | {x}^{2} - 4 |$
$= | x - 2 | \cdot | x + 2 |$

$< \delta \cdot \frac{\epsilon}{\delta}$

$= \epsilon$

This then proves that ${\lim}_{x \to 2} {x}^{2} = 4$