[출제의도] 함수의 극한값을 구할 수 있는가?
lim x → 0 ln ( x + 1 ) x + 4 − 2 \displaystyle \lim\limits _{x \rightarrow 0} {} \frac{\ln \left( x + 1 \right)}{\sqrt{x + 4} - 2} x → 0 lim x + 4 − 2 ln ( x + 1 ) = lim x → 0 { ln ( x + 1 ) × 1 x + 4 − 2 } \displaystyle = \lim\limits _{x \rightarrow 0} {} \left\{ \ln \left( x + 1 \right) \times \frac{1}{\sqrt{x + 4} - 2} \right\} = x → 0 lim { ln ( x + 1 ) × x + 4 − 2 1 } = lim x → 0 { ln ( x + 1 ) x × x x + 4 − 2 } \displaystyle = \lim\limits _{x \rightarrow 0} {} \left\{ \frac{\ln \left( x + 1 \right)}{x} \times \frac{x}{\sqrt{x + 4} - 2} \right\} = x → 0 lim { x ln ( x + 1 ) × x + 4 − 2 x } = lim x → 0 { ln ( x + 1 ) x \displaystyle = \lim\limits _{x \rightarrow 0} {} \left\{ \frac{\ln \left( x + 1 \right)}{x} \right. = x → 0 lim { x ln ( x + 1 ) × x ( x + 4 + 2 ) ( x + 4 − 2 ) ( x + 4 + 2 ) } \displaystyle \times \frac{x \left( \sqrt{x + 4} + 2 \right)}{\left( \sqrt{x + 4} - 2 \right) \left( \sqrt{x + 4} + 2 \right)} \} × ( x + 4 − 2 ) ( x + 4 + 2 ) x ( x + 4 + 2 ) } = lim x → 0 { ln ( x + 1 ) x × ( x + 4 + 2 ) } \displaystyle = \lim\limits _{x \rightarrow 0} {} \left\{ \frac{\ln \left( x + 1 \right)}{x} \times \left( \sqrt{x + 4} + 2 \right) \right\} = x → 0 lim { x ln ( x + 1 ) × ( x + 4 + 2 ) } = 1 × ( 2 + 2 ) = 1 \times \left( 2 + 2 \right) = 1 × ( 2 + 2 ) = 4 = 4 = 4