a₁=lim(x→+∞)[(x-1)e^(½π+arctanx)]/x
=lim(x→+∞)[e^(½π+arctanx)-e^(½π+arctanx)/x]
=e^π 拆开的每一个部分都有极限,可以拆开求极限
a₂=lim(x→-∞)[(x-1)e^(½π+arctanx)]/x
=lim(x→-∞)[e^(½π+arctanx)-e^(½π+arctanx)/x]
=1
b₁=lim(x→+∞)[(x-1)e^(½π+arctanx)-e^π·x]
=lim(x→+∞){[xe^(½π+arctanx)-e^π·x]-e^(½π+arctanx)}
=lim(x→+∞){[e^(½π+arctanx)-e^π)]/(1/x)}-e^π
=lim(x→+∞){[e^(½π+arctanx)/[1/(1+x²)]}/(-1/x²)}-e^π
=lim(x→+∞)[-(1+x²)e^(½π+arctanx)/x²]-e^π
=-2e^π
b₂=lim(x→-∞)[(x-1)e^(½π+arctanx)-x]
=lim(x→-∞){[xe^(½π+arctanx)-x]-e^(½π+arctanx)}
=lim(x→-∞){[e^(½π+arctanx)-1)]/(1/x)}-1
=lim(x→-∞){[e^(½π+arctanx)/[1/(1+x²)]}/(-1/x²)}-1
=lim(x→-∞)[-(1+x²)e^(½π+arctanx)/x²]-1
=-2