변곡점 y좌표 극한

$y = \cos ^{n} x$ $y' = - n \cos ^{n-1} x \sin x$ $y'' = n ( n - 1 ) \cos ^{n-2} x \sin ^{2} x - n \cos ^{n-1} x \cos x$ $= n ( n - 1 ) \cos ^{n-2} x ( 1 - \cos ^{2} x ) - n \cos ^{n-1} x \cos x$ $= n ( n - 1 ) \cos ^{n-2} x - n ( n - 1 ) \cos ^{n} x - n \cos ^{n} x$ $= n ( n - 1 ) \cos ^{n-2} x - n ^{2} \cos ^{n} x$ $= \cos ^{n-2} x ( n ^{2} - n - n ^{2} \cos ^{2} x )$ $= \cos ^{n-2} x ( n ^{2} \sin ^{2} x - n )$ $\displaystyle 0 < x < \frac{\pi}{2}$에서 $\cos x \neq 0$이므로 $y'' = 0$에서 $n ^{2} \sin ^{2} x - n = 0$, $\displaystyle \sin ^{2} x = \frac{1}{n}$ 꼭지점의 좌표를 $( b _{n} , a _{n} )$이라 하면 $\displaystyle \sin ^{2} b _{n} = \frac{1}{n}$이므로 $\displaystyle a _{n} = \cos ^{n} b _{n} = ( \cos ^{2} b _{n} ) ^{\frac{n}{2}} = \left( 1 - \frac{1}{n} \right) ^{\frac{n}{2}}$ ∴ $\displaystyle \lim\limits _{n\rightarrow \infty} a _{n} = \lim\limits _{n\rightarrow \infty} \left( 1 - \frac{1}{n} \right) ^{\frac{n}{2}}$ $\displaystyle = \lim\limits _{n\rightarrow \infty} \left( 1 - \frac{1}{n} \right) ^{-n} \} ^{- \frac{1}{2}}$ $\displaystyle = e ^{- \frac{1}{2}}$ $\displaystyle = \frac{1}{\sqrt{e}}$