Let $f(x)=\sqrt{x+1}-\sqrt{x}$ and $x=0.12345\cdot 10^5$ if the numbers are stored in decimal mantissa with $6$ significant digits we will get
$$f(0.12345\cdot 10^5)=\sqrt{0.12346\cdot 10^5}-\sqrt{0.12345\cdot 10^5}=\\=0.111113\cdot 10^3-0.111108\cdot 10^3=0.5\cdot 10^{-2}$$
So the relative error is $$\frac{0.5\cdot 10^{-3}}{0.5\cdot 10^{-2}}=10\%$$
How did we get to this relative error? shouldn't it be $$\delta=\frac{\mid f(x_1)-f(x_2)\mid}{\mid f(x_1)\mid}$$
Where $f(x^1)$ is the exact number and $f(x_2)$ is the approximation
The answer you get is $0.5\cdot 10^{-2}$. The exact answer is closer to $0.45\cdot10^{-2}$. Thus the relative error is $$ \frac{|0.45\cdot 10^{-2}-0.5\cdot10^{-2}|}{|0.45\cdot 10^{-2}|}\\=\frac{0.5\cdot10^{-3}}{0.45\cdot10^{-2}}=11\% $$