극한으로 정의된 함수 연속

$\displaystyle f \left( x \right) = \lim\limits _{n \rightarrow \infty} {\frac{ax ^{n + 2} + 7 x}{3 x ^{n} + 2}}$ (ⅰ) $0 < x < 1$일 때 $\displaystyle \lim\limits _{n \rightarrow \infty} {x ^{n} = 0}$이므로 $\displaystyle f \left( x \right) = \frac{7}{2} x$ (ⅱ) $x = 1$일 때 $\displaystyle \lim\limits _{n \rightarrow \infty} {x ^{n}} = 1$이므로 $\displaystyle f \left( 1 \right) = \frac{a + 7}{5}$ (ⅲ) $x > 1$일 때 $\displaystyle \lim\limits _{n \rightarrow \infty} {x ^{n} = \infty}$이므로 $f \left( x \right)$$\displaystyle = \lim\limits _{n \rightarrow \infty} {\frac{ax ^{n + 2} + 7 x}{3 x ^{n} + 2}} = \lim\limits _{n \rightarrow \infty} {\frac{ax ^{2} + \frac{7}{x ^{n - 1}}}{3 + \frac{2}{x ^{n}}}}$ $\displaystyle = \frac{a}{3} x ^{2}$ (ⅰ), (ⅱ), (ⅲ)에 의해 $\displaystyle f \left( x \right) = {\begin{cases} \frac{7}{2} x _{{} _{}} & \left( 0 < x < 1 \right) _{{} _{}} \\ \frac{a + 7}{5 _{{} _{}} ^{{} ^{}}} & \left( x = 1 \right) _{{} _{}} ^{{} _{{}_{}} ^{}} \\ \frac{a}{3} x ^{2 ^{{} ^{}}} & \left( x > 1 \right) ^{{} ^{}} \end{cases}}$ 함수 $f \left( x \right)$가 $x = 1$에서 연속이므로 $\displaystyle \lim\limits _{x \rightarrow 1 -} {\frac{7}{2} x} = \lim\limits _{x \rightarrow 1 +} {\frac{a}{3} x ^{2} = f \left( 1 \right)}$ $\displaystyle \frac{7}{2} = \frac{a}{3} = \frac{a + 7}{5}$, $\displaystyle a = \frac{21}{2}$ $\therefore$ $20 a = 210$