Proving a Function is continuous on an interval.
For the function $f(x) = \frac {1}{\sqrt{x}}$ Show the function is continuous on (0, $\infty$)
How do I approach/do this question?
$\endgroup$ 22 Answers
$\begingroup$Let $A\subseteq\mathbb{R}$ and let $f:A\to\mathbb{R}$. Let $c\in A$.
Then $f(x)$ is continuous at $c$ iff for every $\varepsilon>0$, $\exists$ $\delta>0$ such that
$|x-c|<\delta\implies |f(x)-f(c)|<\varepsilon.$
In your case, $f(x)=\frac{1}{\sqrt{x}}$ ,
So $|f(x)-f(c)| = |\frac{1}{\sqrt{x}} - \frac{1}{\sqrt{c}}|$
Try to prove that this value is bounded , given $|x-c|<\delta$.
Here is a similar problem : Continuous function proof by definition
$\endgroup$ $\begingroup$Try following this:
$ f(x) $ is continuous on interval $ (a,b) $
if
$$ \lim\limits_{x \rightarrow a^{+}} f(x) = f(a) $$
and
$$ \lim\limits_{x \rightarrow b^{-}} f(x) = f(b) $$
It is also important to note that if $f$ is differentiable on some interval $(a,b)$ then $f$ is continuous on that interval.
$\endgroup$
