$\begin{align}  & \sqrt{9+{{x}^{2}}}={{(9+{{x}^{2}})}^{\tfrac{1}{2}}}={{\left[ 9(1+\tfrac{{{x}^{2}}}{9}) \right]}^{\tfrac{1}{2}}}={{9}^{\tfrac{1}{2}}}{{(1+\tfrac{{{x}^{2}}}{9})}^{\tfrac{1}{2}}} \\ & \sqrt{9+{{x}^{2}}}=3{{(1+\tfrac{x}{9})}^{\tfrac{1}{2}}} \\ & \sqrt{9+{{x}^{2}}}=3\left[ 1+\frac{1}{2}{{\left( {{\frac{x}{9}}^{2}} \right)}^{2}}+\frac{\tfrac{1}{2}(\tfrac{1}{2}-1)}{2!}{{\left( \frac{{{x}^{2}}}{9} \right)}^{2}}+\frac{\tfrac{1}{2}(\tfrac{1}{2}-1)(\tfrac{1}{2}-2)}{3!}{{\left( \frac{{{x}^{2}}}{9} \right)}^{3}}+--- \right] \\ & \sqrt{9+{{x}^{2}}}=3\left[ 1+\frac{{{x}^{2}}}{18}+\frac{\tfrac{1}{2}(-\tfrac{1}{2})}{2!}\left( \frac{{{x}^{4}}}{81} \right)+\frac{\tfrac{1}{2}(-\tfrac{1}{2})(-\tfrac{3}{2})}{6}\left( \frac{{{x}^{6}}}{729} \right)+--- \right] \\ & \sqrt{9+{{x}^{2}}}=3\left[ 1+\frac{{{x}^{2}}}{18}-\frac{{{x}^{4}}}{648}+\frac{{{x}^{6}}}{\text{11664}} \right] \\ & \text{The range is valid within }-1<\tfrac{{{x}^{2}}}{9}<1 \\\end{align}$