此题打错了,原题应该是：求极限:lim(x→+∞）(2/π arctanx)^x
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).
