解法一：∵∫2arctantdt=2xarctanx-2∫tdt/(1+t²) (应用分部积分法)
=2xarctanx-ln(1+x²)
lim(x->+∞)[ln(1+x²)/x]=lim(x->+∞)[2x/(1+x²)] (∞/∞型极限,应用罗比达法则)
=lim(x->+∞)[(2/x)/(1+1/x²)]
=0
∴原式=lim(x->+∞)[(2xarctanx-ln(1+x²))/√(1+x²)]
=lim(x->+∞)[(2arctanx-ln(1+x²)/x)/√(1+1/x²)] (分子分母同除x)
=[2(π/2)-0]/√(1+0)
=π；
解法二：原式=lim(x->+∞)[2arctanx/(x/√(1+x²))] (∞/∞型极限,应用罗比达法则)
=2[lim(x->+∞)(arctanx)]*{lim(x->+∞)[√(1+1/x²]}
=2(π/2)*√(1+0)
=π.