Question

求解这道极限和证明定积分题

Answer 1

第一题我写了两种解法。
解法一：原式=e^{lim(x->+∞)[x(ln(arctanx)+ln(2/π))]} (应用初等函数的连续性和对数性质)
=e^{lim(x->+∞)[(ln(arctanx)+ln(2/π))/(1/x)]} 
=e^{lim(x->+∞)[((1/arctanx)(1/(1+x²)))/(-1/x²)]} (0/0型极限,应用罗比达法则)
=e^{lim(x->+∞)[(1/arctanx)(-1/(1+1/x²))]}
=e^[(1/(π/2))(-1/(1+0))]
=e^(-2/π)；
解法二：原式=lim(x->+∞){[(1+(2arctanx-π)/π)^(π/(2arctanx-π))]^[x(2arctanx-π)/π]}
={lim(x->+∞)[(1+(2arctanx-π)/π)^(π/(2arctanx-π))]}^{lim(x->+∞)[x(2arctanx-π)/π]}
=e^{lim(x->+∞)[x(2arctanx-π)/π]} (应用重要极限lim(z->0)[(1+z)^(1/z)]=e)
=e^{lim(x->+∞)[(2arctanx-π)/(π/x)]} 
=e^{lim(x->+∞)[(2/(1+x²))/(-π/x²)]} (0/0型极限,应用罗比达法则)
=e^{lim(x->+∞)[(-2/π)(1/(1+1/x²))]}
=e^[(-2/π)(1/(1+0))]
=e^(-2/π).

第二题
证明：设F(x)=∫(0,x)f(t)dt
F(-x)=∫(0,-x)f(t)dt,对此积分,代换t=-y,当t=0时，y=0，当t=-x时，-y=-x，y=x，此时积分上下限变为(0,x)
代入得：
F(-x)=∫(0,-x)f(t)dt
=∫(0,x)[-f(-y)]dy=∫(0,x)[-f(-t)]dt
如果f(t)是连续的奇函数,那么：f(-t)=-f(t) ,F(-x)=∫(0,x)[f(t)]dt=F(x),F(x)为偶函数.
如果f(t)是连续的偶函数,那么：f(-t)=f(t) ,F(-x)=∫(0,x)[-f(t)]dt=-F(x),F(x)为奇函数.