극한 함수 합성 연속

$\displaystyle f ( x ) = \lim\limits _{h \rightarrow \infty} {\frac{x ^{2n+1}}{x ^{2n} + 1}}$에서 $\mathrm{i} ) \left| x \right| > 1 , f ( x ) = x$ $\mathrm{ii} ) \left| {x} \right| < 1 , f ( x ) = 0$ $\displaystyle \mathrm{iii} ) x = 1 , f ( x ) = \frac{1}{2}$ $\displaystyle \mathrm{iv} ) x = - 1 , f ( x ) = - \frac{1}{2}$ $\therefore x = \pm 1$에서 불연속 $y = g ( f ( x ) )$에서 $x = \pm 1$에서 연속이 되려면 $\displaystyle \lim\limits _{x \rightarrow 1} {g ( f ( x ) )} = g ( f ( 1 ) )$, $\displaystyle \lim\limits _{x \rightarrow - 1} {g ( f ( x ) )} = g ( f ( - 1 ) )$ 위 식을 만족하려면 좌극한값, 우극한값, 함수값이 같아야 하므로 $\displaystyle \lim\limits _{x \rightarrow 1 + 0} {g ( f ( x ) )} = \lim\limits _{x \rightarrow 1 + 0} {} g ( x ) = g ( 1 )$ $\displaystyle \lim\limits _{x \rightarrow 1 - 0} {g ( f ( x ) )} = \lim\limits _{x \rightarrow 0} {} g ( x ) = g ( 0 )$ $\displaystyle g ( f ( 1 ) ) = g \left( \frac{1}{2} \right)$ $\displaystyle g ( 1 ) = g ( 0 ) = g \left( \frac{1}{2} \right)$ 이와같은 방법으로 $\displaystyle \lim\limits _{x \rightarrow - 1 + 0} {g ( f ( x ) )} = \lim\limits _{x \rightarrow 0} {} g ( x ) = g ( 0 )$ $\displaystyle \lim\limits _{x \rightarrow - 1 - 0} {g ( f ( x ) )} = \lim\limits _{x \rightarrow - 1 - 0} g ( x ) = g ( - 1 )$ $\displaystyle g ( f ( - 1 ) ) = g \left( - \frac{1}{2} \right)$ $\displaystyle g ( 0 ) = g ( - 1 ) = g \left( - \frac{1}{2} \right)$을 만족시키는 $y = g ( x )$의 개형은 ④번