Let $f(x) = \frac1{\sqrt x}$ for $x\in(0,\infty)$. Given $\varepsilon>0$ and $x_0\in(0,\infty)$, show there exists $\delta>0$ such that $$|x-x_0|<\delta$$ implies that $$|f(x)-f(x_0)| <\varepsilon.$$
Now $$|f(x)-f(x_0)| = \left|\frac 1{\sqrt x} - \frac1{\sqrt{x_0}}\right| = \frac{|x_0-x|}{|x_0\sqrt x + x\sqrt{x_0}|}.$$
This was done by putting them under one denominator, and multiplying top and bottom by the conjugate of the top.
Where do I go from here?
First choose $\delta=\frac{x_0}{2}$. Then, for this value of $\delta$, $x$ is in the interval $\frac{x_0}{2}<x<\frac{3x_0}{2}$. Then, we have the following
$$\begin{align} \left|\frac{1}{\sqrt{x}} -\frac{1}{\sqrt{x_0}}\right|&=\left|\frac{\sqrt{x}-\sqrt{x_0}}{\sqrt{xx_0}} \right|\\ &=\left|\frac{x-x_0}{\sqrt{xx_0}(\sqrt{x}+\sqrt{x_0})} \right|\\ &\le \frac{\left|x-x_0\right|}{\sqrt{(\frac12x_0)x_0}\;\left(\sqrt{\frac12x_0}+\sqrt{x_0}\right)} \\ &<\frac{\left|x-x_0\right|}{x_0^{3/2}}\\ &< \epsilon \end{align}$$
when $|x-x_0|<\delta =\min (x_0/2,x_0^{3/2}\epsilon)$.