Question

高数求极限，谢谢

如图

Answer 1

等价无穷小代换结合洛必达法则可以求出结果。

Answer 2

这是求 y = (x-1)e^(π/2+arctanx) 的渐近线问题。
无水平、垂直渐近线。设斜渐近线 y = kx+b， 则
k = lim<x→∞>y/x = lim<x→∞>(x-1)e^(π/2+arctanx)/x
k1 = lim<x→+∞>(x-1)e^(π/2+arctanx)/x
= lim<x→+∞>(1-1/x)e^(π/2+arctanx)= e^π;
k2 = lim<x→-∞>(x-1)e^(π/2+arctanx)/x
= lim<x→-∞>(1-1/x)e^(π/2+arctanx)= 1.
b1 = lim<x→+∞>(y-k1x)
= lim<x→+∞>[(x-1)e^(π/2+arctanx)-(e^π)x] = -e^π
b2 = lim<x→-∞>(y-k1x)
= lim<x→-∞>[(x-1)e^(π/2+arctanx)-x] = -1
两条斜渐近线分别为 y = (e^π)(x-1), y = x-1