극한과 미정계수

$\displaystyle \lim\limits _{x \rightarrow 3} \frac{\sqrt{x + a} - b}{x - 3} = \frac{1}{4}$에서 $x \rightarrow 3$일 때 (분모) → 0이므로 (분자) → 0이어야 한다. 즉, $\displaystyle \lim\limits _{x \rightarrow 3} ( \sqrt{x + a} - b ) = 0$에서 $\displaystyle \sqrt{3 + a} - b = 0$ $\displaystyle \therefore b = \sqrt{3 + a}$ $\displaystyle \lim\limits _{x \rightarrow 3} \frac{\sqrt{x + a} - \sqrt{3 + a}}{x - 3}$ $\displaystyle = \lim\limits _{x \rightarrow 3} \frac{x - 3}{( x - 3 ) \left( {\sqrt{x + a} + \sqrt{3 + a}} \right)}$ $\displaystyle = \lim\limits _{x \rightarrow 3} \frac{1}{\sqrt{x + a} + \sqrt{3 + a}}$$\displaystyle = \frac{1}{2 \sqrt{3 + a}} = \frac{1}{4}$ 즉, $\displaystyle 2 \sqrt{3 + a} = 4$에서 $\displaystyle \sqrt{3 + a} = 2$ ∴ $a = 1$ 따라서, $\displaystyle b = \sqrt{3 + 1} = 2$이고 $a + b = 3$