로그 절댓값 미분 극한

$F ( x ) = \ln \left| f ( x ) \right|$를 미분하면 $\displaystyle F' ( x ) = \frac{f' ( x )}{f ( x )}$ $\displaystyle \lim\limits _{x \rightarrow 1} ( x - 1 ) F' ( x ) = \lim\limits _{x \rightarrow 1} \frac{( x - 1 ) f' ( x )}{f ( x ) = 3}$에서 $f ( 1 ) = 0$이다. $f ( x ) = ( x - 1 ) q ( x )$라 하면 $f' ( x ) = q ( x ) + ( x - 1 ) q' ( x )$이므로 $\displaystyle \lim\limits _{x \rightarrow 1} \frac{( x - 1 ) \left\{ Q ( x ) + ( x - 1 ) Q' ( x ) \right\}}{( x - 1 ) Q ( x )} = 1 \ne 3$이다. $f ( x ) = ( x - 1 ) ^{k} h ( x ) ( k \geq 2 \text{인} \text{자연수} )$라 하면 $f' ( x ) = k ( x - 1 ) ^{k - 1} h ( x ) + ( x - 1 ) ^{k} h' ( x )$이므로 $\displaystyle \lim\limits _{x \rightarrow 1} \frac{( x - 1 ) \left\{ k ( x - 1 ) ^{k - 1} h ( x ) + ( x - 1 ) ^{k} h' ( x ) \right\}}{( x - 1 ) ^{k} h ( x )}$ $\displaystyle = \lim\limits _{x \rightarrow 1} \frac{k ( x - 1 ) ^{k} h ( x ) + ( x - 1 ) ^{k + 1} h' ( x )}{( x - 1 ) ^{k} h ( x )}$ $= k$ 이므로 조건으로부터 $k = 3$이다. 따라서 $f ( x ) = ( x - 1 ) ^{3} ( x - a )$의 꼴이 된다. $\displaystyle \lim\limits _{x \rightarrow 0} \frac{F' ( x )}{G' ( x )}$$\displaystyle = \lim\limits _{x \rightarrow 0} \frac{f' ( x ) \times g ( x ) \sin x}{f ( x ) \left\{ g' ( x ) \sin x + g ( x ) \cos x \right\}} = \frac{1}{4} \cdots \text{①}$ 에서 $x \rightarrow 0$일 때 $( \text{분자} ) \rightarrow 0$이므로 $( \text{분모} ) \rightarrow 0$이어야 한다. 따라서 $f ( 0 ) g ( 0 ) = 0$이다.

$i )$ $f ( 0 ) = a \not = 0$일 때 $g ( 0 ) = 0$ $f ( x ) = ( x - 1 ) ^{3} ( x - a )$이고 $f' ( x ) = 3 ( x - 1 ) ^{2} ( x - a ) + ( x - 1 ) ^{3} = ( x - 1 ) ^{2} ( 4 x - 3 a - 1 )$이고 $\displaystyle \frac{f' ( x )}{f ( x )} = \frac{{( x - 1 ) ^{2} ( 4 x - 3 a - 1 )}}{( x - 1 ) ^{3} ( x - a )} = \frac{4 x - 3 a - 1}{( x - 1 ) ( x - a )}$ $\displaystyle \lim\limits _{x \rightarrow 0} \frac{F' ( x )}{G' ( x )}$$\displaystyle = \lim\limits _{x \rightarrow 0} \frac{( 4 x - 3 a - 1 ) g ( x ) \sin x}{( x - 1 ) ( x - a ) \left\{ g' ( x ) \sin x + g ( x ) \cos x \right\}}$ $\displaystyle = \lim\limits _{x \rightarrow 0} \frac{4 x - 3 a - 1}{( x - 1 ) ( x - a )} \cdot \frac{\sin x}{x} \cdot \frac{xg ( x )}{g' ( x ) \sin x + g ( x ) \cos x} = 0 \not = \frac{1}{4}$ 이 되어 모순이다. 따라서 $f ( 0 ) = 0$이다.

$ii )$ $f ( 0 ) = 0$일 때 $f ( x ) = ( x - 1 ) ^{3} x$이다. $x$에 대해 미분하면 $f' ( x ) = 3 ( x - 1 ) ^{2} x + ( x - 1 ) ^{3}$이므로 $\displaystyle \frac{f' ( x )}{f ( x )} = \frac{3 ( x - 1 ) ^{2} x + ( x - 1 ) ^{3}}{( x - 1 ) ^{3} x} = \frac{4 x - 1}{x ( x - 1 )}$이다. 위의 식을 $\text{①}$에 대입하면 $\displaystyle \lim\limits _{x \rightarrow 0} \frac{F' ( x )}{G' ( x )}$$\displaystyle = \lim\limits _{x \rightarrow 0} \frac{( 4 x - 1 ) g ( x ) \sin x}{x ( x - 1 ) \left\{ g' ( x ) \sin x + g ( x ) \cos x \right\}}$ $\displaystyle = \lim\limits _{x \rightarrow 0} \frac{4 x - 1}{x - 1} \cdot \frac{\sin x}{x} \cdot \frac{g ( x )}{g' ( x ) \sin x + g ( x ) \cos x} = \frac{1}{4}$ 에서 $g ( 0 ) = 0$이다. $g ( x ) = xk ( x )$라 하면 $g' ( x ) = k ( x ) + xk' ( x )$이므로 $\displaystyle \lim\limits _{x \rightarrow 0} \frac{F' ( x )}{G' ( x )}$ $\displaystyle = \lim\limits _{x \rightarrow 0} \frac{4 x - 1}{x - 1} \cdot \frac{\sin x}{x} \cdot \frac{x k ( x )}{\left\{ k ( x ) + xk' ( x ) \right\} \sin x + x k ( x ) \cos x} = \frac{1}{4}$ 에서 $k ( 0 ) = 0$이다. 따라서 $g ( x ) = x ^{m} p ( x ) ( m \geq 2 \text{인} \text{자연수} )$라 하면 $g' ( x ) = m x ^{m - 1} p ( x ) + x ^{m} p' ( x )$이므로 $\displaystyle \lim\limits _{x \rightarrow 0} \frac{F' ( x )}{G' ( x )}$$\displaystyle = \lim\limits _{x \rightarrow 0} \frac{4 x - 1}{x - 1} \cdot \frac{\sin x}{x} \cdot \frac{x ^{m} p ( x )}{\left\{ mx ^{m - 1} p ( x ) + x ^{m} p' ( x ) \right\} \sin x + x ^{m} p ( x ) \cos x}$ $\displaystyle = \lim\limits _{x \rightarrow 0} \frac{4 x - 1}{x - 1} \cdot \frac{\sin x}{x} \cdot \frac{x ^{m - 1} p ( x )}{\left\{ mx ^{m - 1} p ( x ) + x ^{m} p' ( x ) \right\} \frac{\sin x}{x} + x ^{m - 1} p ( x ) \cos x}$ $\displaystyle = 1 \cdot 1 \cdot \frac{1}{m + 1}$$\displaystyle = \frac{1}{m + 1}$ 이므로 조건으로부터 $m = 3$이다. 따라서 $g ( x ) = x ^{3}$이다. 그러므로 $f ( 3 ) + g ( 3 )$$= 3 \cdot 2 ^{3} + 3 ^{3} = 51$이다.