# How do you use the epsilon-delta definition of continuity to prove f(x) = x^2 is continuous?

Dec 15, 2016

For any $\epsilon > 0$ we chose $\delta$ such that:

$\delta \left(2 | \overline{x} | + \delta\right) < \epsilon$

#### Explanation:

By definition, $f \left(x\right)$ is continuous for $x = \overline{x}$, if for any real number $\epsilon > 0$ we can find a real number $\delta > 0$ such that:

$x \in \left(\overline{x} - \delta , \overline{x} + \delta\right) \implies | f \left(x\right) - f \left(\overline{x}\right) | < \epsilon$

Now, for $x \in \left(\overline{x} - \delta , \overline{x} + \delta\right)$, we have that:

$| x - \overline{x} | < \delta$,
$| x | < | \overline{x} | + \delta$

and then:

$| f \left(x\right) - f \left(\overline{x}\right) | = | {x}^{2} - {\overline{x}}^{2} | = | \left(x - \overline{x}\right) \left(x + \overline{x}\right) | = | x - \overline{x} | | x + \overline{x} | \le \delta \left(| x | + | \overline{x} |\right) \le \delta \left(2 | \overline{x} | + \delta\right)$

So, for any $\epsilon > 0$ if we chose $\delta$ such that:

$\delta \left(2 | \overline{x} | + \delta\right) < \epsilon$

The condition of continuity is satisfied.