find the value of

$x=\sqrt{13}+2 \sqrt{3} \quad \ldots .(1)$

$\Rightarrow \frac{1}{x}=\frac{1}{\sqrt{13}+2 \sqrt{3}}$

$\Rightarrow \frac{1}{x}=\frac{1}{\sqrt{13}+2 \sqrt{3}} \times \frac{\sqrt{13}-2 \sqrt{3}}{\sqrt{13}-2 \sqrt{3}}$

$\Rightarrow \frac{1}{x}=\frac{\sqrt{13}-2 \sqrt{3}}{(\sqrt{13})^{2}-(2 \sqrt{3})^{2}}$

$\Rightarrow \frac{1}{x}=\frac{\sqrt{13}-2 \sqrt{3}}{13-12}$

$\Rightarrow \frac{1}{x}=\sqrt{13}-2 \sqrt{3}$          ..........(2)

Subtracting (2) from (1), we get

$x-\frac{1}{x}=(\sqrt{13}+2 \sqrt{3})-(\sqrt{13}-2 \sqrt{3})$

$\Rightarrow x-\frac{1}{x}=\sqrt{13}+2 \sqrt{3}-\sqrt{13}+2 \sqrt{3}$

$\Rightarrow x-\frac{1}{x}=4 \sqrt{3}$

Thus, the value of $x-\frac{1}{x}$ is $4 \sqrt{3}$.