Proving $\frac{1}{\sqrt x} $is discontinuous at 0 The function at hand is $f(x) = \frac{1}{\sqrt x} $
I know that is order to prove discontinuity, I have to show the following:
$\forall \delta \gt 0, \exists \varepsilon \gt 0 ~~\text{and} ~~ \exists x ~~ \text{such} ~\text{that} ~~|x - 0| \lt \delta ~~ \text{implies} ~~ |f(x) - f(0)| \geq \varepsilon$
but the function is not defined at $x = 0$, so I can't develop $f(0)$.
I also tried finding a sequence that converges to $0$ but it's image does not converge to the image of $0$, but I'm faced with the same problem there as well.
grateful for any help. 

I think the question lying under the question (because like the comments said, the question makes no sense since $f$ is not even defined at $x=0$) is:
Is it possible to expand by continuity $f(x)=\frac 1{\sqrt x}$ at $x=0$.
And the answer is no because $$\lim_{x\to 0^+}f(x)=+\infty$$
so $f$ does not have a finite limit at $x=0^+$.
