,

จาก $a_n= \sqrt{n^2+16n+3} - \sqrt{n^2+2}$ เมื่อลองดู $n\rightarrow\infty$   จะได้ $\infty - \infty$ ซึ่งไม่สามารถบอกได้ว่ามีค่าเท่าใด

ดังนั้นเราจะจัดรูป $a_n$ ให้ โดยการใช้ คอนจูเกจ จะได้

\begin{eqnarray*}
a_n&=& \sqrt{n^2+16n+3} - \sqrt{n^2+2}\\
&=& \sqrt{n^2+16n+3} - \sqrt{n^2+2}\cdot\left(\frac{ \sqrt{n^2+16n+3} + \sqrt{n^2+2}}{ \sqrt{n^2+16n+3} + \sqrt{n^2+2}}\right)\\
&=&\frac{\left( \sqrt{n^2+16n+3} - \sqrt{n^2+2}\right)\left( \sqrt{n^2+16n+3} + \sqrt{n^2+2}\right)}{ \sqrt{n^2+16n+3} + \sqrt{n^2+2}}\\
&=&\frac{\left(n^2+16n+3\right)-\left(n^2+2\right)}{ \sqrt{n^2+16n+3} + \sqrt{n^2+2}}\\
&=&\frac{16n+1}{ \sqrt{n^2+16n+3} + \sqrt{n^2+2}}
\end{eqnarray*}

,

\begin{eqnarray*}
\lim_{n\rightarrow\infty}\sqrt[3]{a_n}&=&\lim_{n\rightarrow\infty}\sqrt[3]{\frac{16n+1}{ \sqrt{n^2+16n+3} + \sqrt{n^2+2}}}\\
&=&\sqrt[3]{\lim_{n\rightarrow\infty}{\frac{16n+1}{ \sqrt{n^2+16n+3} + \sqrt{n^2+2}}}}\\
&=&\sqrt[3]{\lim_{n\rightarrow\infty}{\frac{\frac{16n+1}{n}}{\frac{ \sqrt{n^2+16n+3} + \sqrt{n^2+2}}{n}}}}\\
&=&\sqrt[3]{\lim_{n\rightarrow\infty}{\frac{16+\frac{1}{n}}{\sqrt{1+\frac{16}{n}+\frac{3}{n^2}}+\sqrt{1+\frac{2}{n^2}}}}}\\
&=&\sqrt[3]{\frac{16+0}{\sqrt{1+0+0}+\sqrt{1+0}}}\\
&=&\sqrt[3]{\frac{16}{2}}\\
&=&\sqrt[3]{8}\\
&=&2
\end{eqnarray*}