How do you prove that the function f(x) = x^(1/2) is continuous at x=1/2?

Oct 31, 2015

There are many ways to prove this - one could use :

1. Epsilon-delta methods
$y = f \left(x\right)$ is continuous at $x = {x}_{0} \iff \forall \epsilon > 0 , \exists \delta > 0$ such that $| x - {x}_{0} | < \delta \implies | f \left(x\right) - f \left({x}_{0}\right) | < \epsilon$

2. Topological methods
y:X->Y;y=f(x) is continuous everywhere (specifically at ${x}_{0} \in X$ iff and only if for every open subset A in Y, the inverse function ${f}^{- 1} \left(A\right)$ is open in X.

3. Functional analysis methods using sequential criterion .
$y = f \left(x\right)$ is continuous at $x = {x}_{0} \iff$ for every sequence $\left({x}_{n}\right)$ converging to ${x}_{0}$, the sequence of function values $\left(f \left({x}_{n}\right)\right)$ converges to $f \left({x}_{0}\right)$.

I shall prove the continuity using the latter definition in this case :

Let $\left({x}_{n}\right) = \left(\frac{1}{2} + \frac{1}{n} ^ 2\right)$ be a sequence in $\mathbb{R}$.
Since $\frac{1}{n} ^ 2 \to 0 \implies \left({x}_{n}\right) \to \frac{1}{2} \in \mathbb{R}$.

But $\left(f \left({x}_{n}\right)\right) = {\left(\frac{1}{2} + \frac{1}{n} ^ 2\right)}^{\frac{1}{2}}$ which converges to ${\left(\frac{1}{2}\right)}^{\frac{1}{2}} = f \left(\frac{1}{2}\right)$

$\therefore f$ is continuous at $x = \frac{1}{2}$.