함수의 극한으로 정의된 함수의 좌극한

$| x | < 1$이면 $\displaystyle f ( x ) = \lim\limits _{n \rightarrow \infty} \frac{x ^{2 n + 1} - 2 x ^{2 n} + 1}{x ^{2 n + 2} + x ^{2 n} + 1}$$= 1$ $| x | > 1$이면 $\displaystyle f ( x ) = \lim\limits _{n \rightarrow \infty} \frac{x ^{2 n + 1} - 2 x ^{2 n} + 1}{x ^{2 n + 2} + x ^{2 n} + 1}$$\displaystyle = \lim\limits _{n \rightarrow \infty} \frac{x - 2 + \frac{1}{x ^{2 n}}}{x ^{2} + 1 + \frac{1}{x ^{2 n}}}$$\displaystyle = \frac{x - 2}{x ^{2} + 1}$ $\displaystyle \lim\limits _{x \rightarrow - 1 -} f ( x ) = \lim\limits _{x \rightarrow - 1 -} \frac{x - 2}{x ^{2} + 1}$$\displaystyle = - \frac{3}{2}$$= a$, $\displaystyle \lim\limits _{x \rightarrow 1 -} f ( x ) = 1$$= b$ $\therefore$ $\displaystyle \frac{b}{a + 2} = \frac{1}{- \frac{3}{2} + 2} = 2$