탄젠트 교점 수열 극한

함수 $\displaystyle y = \frac{\sqrt{x}}{10}$와 함수 $y = \tan x$의 그래프가 만나는 점의 $x$좌표가 $a _{n}$이므로 아래 그림에서처럼

$\displaystyle \frac{\sqrt{x}}{10} = \tan x$에서 $\displaystyle \left( n - 2 \right) \pi \leq x < \left( n - 2 \right) \pi + \frac{\pi}{2}$의 근을 $x = \left( n - 2 \right) \pi + \alpha _{n}$이라 하자. $\displaystyle \left( \begin{array}{l} \begin{matrix} \\ \end{matrix} \end{array} \right.$단, $\displaystyle 0 \leq \alpha _{n} < \frac{\pi}{2} )$ $a _{1} = \alpha _{1} = 0$, $a _{2} = \alpha _{2}$, $a _{3} = \pi + \alpha _{3}$, $a _{4} = 2 \pi + \alpha _{4}$, $\cdots$ $a _{n} = \left( n - 2 \right) \pi + \alpha _{n}$ $\displaystyle \left( \begin{array}{l} \begin{matrix} \\ \end{matrix} \end{array} \right.$단, $\displaystyle 0 \leq \alpha _{n} < \frac{\pi}{2} )$ 이므로 $a _{n+1} = \left( n - 1 \right) \pi + \alpha _{n+1}$ $\alpha _{n} < \alpha _{n+1}$이고, $\displaystyle \lim\limits _{n \rightarrow \infty} {\alpha _{n}} = \frac{\pi}{2}$이므로 $\displaystyle \lim\limits _{n \rightarrow \infty} {\left( \alpha _{n+1} - \alpha _{n} \right)} = 0$ 이고 $\displaystyle \lim\limits _{n \rightarrow \infty} {\frac{a _{n+1}}{a _{n}}}$$\displaystyle = \lim\limits _{n \rightarrow \infty} {\frac{\left( n - 1 \right) \pi + \alpha _{n+1}}{\left( n - 2 \right) \pi + \alpha _{n}}}$$= 1$ 이므로 $\tan \left( a _{n+1} - a _{n} \right)$$\displaystyle = \frac{\tan a _{n+1} - \tan a _{n}}{1 + a _{n+1} a _{n}}$ $\displaystyle = \frac{\frac{\sqrt{a _{n+1}} - \sqrt{a _{n}}}{10}}{1 + \frac{\sqrt{a _{n+1} a _{n}}}{100}}$ $\displaystyle = \frac{10 \left( \sqrt{a _{n+1}} - \sqrt{a _{n}} \right)}{100 + \sqrt{a _{n+1} a _{n}}}$ $\displaystyle = \frac{10 \left( a _{n+1} - a _{n} \right)}{\left( 100 + \sqrt{a _{n+1} a _{n}} \right) \left( \sqrt{a _{n+1}} + \sqrt{a _{n}} \right)}$ $\displaystyle = \frac{10 \left( \pi + \alpha _{n+1} - \alpha _{n} \right)}{\left( 100 + \sqrt{a _{n+1} a _{n}} \right) \left( \sqrt{a _{n+1}} + \sqrt{a _{n}} \right)}$ $\displaystyle a _{n} \sqrt{a _{n}} \times \tan \left( a _{n+1} - a _{n} \right)$ $\displaystyle = \frac{10 \left( \pi + \alpha _{n+1} - \alpha _{n} \right)}{\left( \frac{100}{a _{n}} + \sqrt{\frac{a _{n+1}}{a _{n}}} \right) \left( \sqrt{\frac{a _{n+1}}{a _{n}}} + 1 \right)}$ 따라서 $\displaystyle \frac{1}{\pi ^{2}} \times \lim\limits _{n \rightarrow \infty} {a _{n} ^{3} \tan ^{2} \left( a _{n+1} - a _{n} \right)}$ $\displaystyle = \frac{1}{\pi ^{2}} \times \lim\limits _{n \rightarrow \infty} {\left( \frac{10 \left( \pi + \alpha _{n+1} - \alpha _{n} \right)}{\left( \frac{100}{a _{n}} + \sqrt{\frac{a _{n+1}}{a _{n}}} \right) \left( \sqrt{\frac{a _{n+1}}{a _{n}}} + 1 \right)} \right) ^{2}}$ $\displaystyle = \frac{1}{\pi ^{2}} \times \left( \frac{10 \pi}{\left( 0 + 1 \right) \left( 1 + 1 \right)} \right) ^{2}$ $\displaystyle = \frac{1}{\pi ^{2}} \times 25 \pi ^{2}$ $= 25$