I tried to find the limit of \begin{equation} ((n+a)(n+b))^{\frac{1}{2}}-n =\frac{a+b}{2}, a>0, b>0 \end{equation} I took the help of rationalization \begin{align} ((n+a)(n+b))^{\frac{1}{2}}-n &=((n+a)(n+b))^{\frac{1}{2}}-n \times \frac{((n+a)(n+b))^{\frac{1}{2}}+n}{((n+a)(n+b))^{\frac{1}{2}}+n}\\ &=\frac{(n+a)(n+b)-n^{2}}{{((n+a)(n+b))}^{\frac{1}{2}}+n}\\ &=\frac{n^{2}[(1+\frac{a}{n})(1+\frac{b}{n})-1]}{n[({(1+\frac{a}{n})(1+\frac{b}{n}))}^{\frac{1}{2}}+1]}\\ &=\frac{n^{2}[(1+\frac{a}{n})(1+\frac{b}{n})-1]}{n[{((1+\frac{a}{n})(1+\frac{b}{n}))}^{\frac{1}{2}}+1]}\\ &=\frac{n[(1+\frac{a}{n})(1+\frac{b}{n})-1]}{{((1+\frac{a}{n})(1+\frac{b}{n}))}^{\frac{1}{2}}+1}\\ \end{align}
After that, I am unable to solve the problem further. Would like to get few valuable tips from you.
