# How do you prove that sqrtx is continuous?

Dec 14, 2016

We need to prove that for any point $a \in \left(0 , \infty\right)$, for every $\epsilon > 0$ there exists a $\delta > 0$ such that

$| x - a | < \delta \implies | \sqrt{x} - \sqrt{a} | < \epsilon$

So, to find a suitable $\delta$, we must look at the inequality $| \sqrt{x} - \sqrt{a} | < \epsilon$. Since we want an expression involving |x−a|, then multiply by the conjugate to remove the square roots, so:

$\setminus \setminus \setminus \setminus \setminus | \sqrt{x} - \sqrt{a} | < \epsilon$
$\therefore | \sqrt{x} - \sqrt{a} | \cdot | \sqrt{x} + \sqrt{a} | < \epsilon \cdot | \sqrt{x} - \sqrt{a} |$
$\therefore | \left(\sqrt{x} - \sqrt{a}\right) \cdot \left(\sqrt{x} + \sqrt{a}\right) | < \epsilon \cdot | \sqrt{x} - \sqrt{a} |$
$\therefore | \sqrt{x} \sqrt{x} + \sqrt{x} \sqrt{a} - \sqrt{x} \sqrt{a} - \sqrt{a} \sqrt{a} | < \epsilon \cdot | \sqrt{x} - \sqrt{a} |$
$\therefore | x - a | < \epsilon \cdot | \sqrt{x} - \sqrt{a} |$ ..... [1]

Now, if you require that |x−a|<1, then it follows that x−a < 1, so:

 \ \ \ \ \ \ a−1 < x < a+1
$\therefore \sqrt{x} < \sqrt{a + 1}$.
$\therefore \sqrt{x} + \sqrt{a} < \sqrt{a + 1} + \sqrt{a}$,

which combined with [1] gives;

 |x−a| < epsilon (sqrt(a+1) + sqrt(a))

So, let $\delta = \min \left(1 , \epsilon \left(\sqrt{a + 1} + \sqrt{a}\right)\right)$.

Hence we have proved that $f \left(x\right) = \sqrt{x}$ is continuous on $\left(0 , \infty\right)$