자연계열 1번

[1-1 풀이]$\displaystyle f(1)=0$이므로 제시문에 의하여 $\displaystyle g(f(1))=1$따라서$\displaystyle \lim_{x\to1}\frac{g(f(x))-1}{x-1}=\lim_{x\to1}\frac{g(f(x))-g(f(1))}{x-1}=(g\circ f)^{\prime}(1)$

여기서 $\displaystyle h(x)=(g\circ f)(x)=g(f(x))$라 두면 $\displaystyle h^{\prime}(x)=g^{\prime}(f(x))f^{\prime}(x)$이므로$\displaystyle h^{\prime}(1)=g^{\prime}(f(1))f^{\prime}(1)=g^{\prime}(0)f^{\prime}(1)$한편 $\displaystyle f^{\prime}(1)=-\frac{\pi}{2}$이고 $\displaystyle g^{\prime}(0)=1$이므로$\displaystyle \lim_{x\to1}\frac{g(f(x))-1}{x-1}=h^{\prime}(1)=-\frac{\pi}{2}$

[1-2 풀이]$\displaystyle \begin{aligned}\lim_{x\to0}\frac{\sin2x}{x(e^x-1)}\int_0^xg(t)dt&=\lim_{x\to0}\left(\frac{\sin2x}{2x}\right)\left(\frac{2x}{e^x-1}\right)\left(\frac1x\int_0^xg(t)dt\right)\\&=\left(\lim_{x\to0}\frac{\sin2x}{2x}\right)\left(\lim_{x\to0}\frac{2x}{e^x-1}\right)\left(\lim_{x\to0}\frac1x\int_0^xg(t)dt\right)\quad\cdots\cdots\end{aligned}$ ㉠이다.$\displaystyle G(x)=\int_0^xg(t)dt$라 하면$\displaystyle \lim_{x\to0}\frac1x\int_0^xg(t)dt=\lim_{x\to0}\frac{G(x)-G(0)}{x-0}=G^{\prime}(0)$

한편, $\displaystyle \lim_{x\to0}\frac{\sin2x}{2x}=1$이고 $\displaystyle \lim_{x\to0}\frac{2x}{e^x-1}=2\lim_{x\to0}\frac{x}{e^x-1}=2\times1=2$이다.

따라서 $\displaystyle G^{\prime}(0)=g(0)=1$이므로 구하는 값은 ㉠에서$\displaystyle \lim_{x\to0}\frac{\sin2x}{x(e^x-1)}\int_0^xg(t)dt=1\times2\times1=2$