극한값으로 상수

[출제의도] 함수의 극한값 계산하기 $\displaystyle \lim\limits _{x \rightarrow - 1} {} \frac{x ^{2} + 4 x + a}{x + 1}$$= b$에서 (분모)$\rightarrow$$0$이고 극한값이 존재하므로 (분자)$\rightarrow$$0$ $\displaystyle \lim\limits _{x \rightarrow - 1} {} \left( x ^{2} + 4 x + a \right)$$= 0$이므로 $1 - 4 + a$$= 0$, $a = 3$ $\displaystyle \lim\limits _{x \rightarrow - 1} {} \frac{x ^{2} + 4 x + 3}{x + 1}$$\displaystyle = \lim\limits _{x \rightarrow - 1} {} \frac{\left( x + 1 \right) \left( x + 3 \right)}{x + 1}$ $\displaystyle = \lim\limits _{x \rightarrow - 1} {} \left( x + 3 \right)$$= \mathit{2}$$= b$ 따라서 $a + b$$= 5$