두 극한이 같을 조건

$\displaystyle \lim\limits _{n \rightarrow \infty} \frac{\left( bn - 1 \right) ^{2}}{\left( b + 6 \right) n ^{2} + 1} = \lim\limits _{n \rightarrow \infty} \frac{\left( b - \frac{1}{n} \right) ^{2}}{b + 6 + \frac{1}{n ^{2}}} = \frac{b ^{2}}{b + 6}$ 이므로 $\displaystyle \lim\limits _{n \rightarrow \infty} \left( \sqrt{an ^{2} + bn} - bn \right)$$\displaystyle = \lim\limits _{n \rightarrow \infty} \frac{\left( a - b ^{2} \right) n ^{2} + bn}{\sqrt{an ^{2} + bn} + bn}$ $\displaystyle = \lim\limits _{n \rightarrow \infty} \frac{\left( a - b ^{2} \right) n + b}{\sqrt{a + \frac{b}{n}} + b} = \frac{b ^{2}}{b + 6}$ 그러므로 $a - b ^{2} = 0$, $a = b ^{2}$이고 $\displaystyle \lim\limits _{n \rightarrow \infty} \frac{b}{\sqrt{b ^{2} + \frac{b}{n}} + b} = \frac{1}{2} = \frac{b ^{2}}{b + 6}$ $2 b ^{2} - b - 6 = \left( 2 b + 3 \right) \left( b - 2 \right) = 0$ $b > 0$이므로 $b = 2$, $a = 4$ 따라서 $a + b = 6$