一年級極限數學題討論

求：
lim(t->1-) sqrt(1-t)(t^(1^2) + t^(2^2) + t^(3^2) + ... +  t^(n^2)  + ...)

令 dx = sqrt(1-t)
t = 1 - dx^2

sqrt(1-t)(t^(1^2) + t^(2^2) + t^(3^2) + ... +  t^(n^2)  + ...) =
dx( (1 - dx^2)^(1^2) + (1 - dx^2)^(2^2) + (1 - dx^2)^(3^2) + ... +  (1 - dx^2)^(n^2) + ...) =
dx( [(1 - dx^2)^ (1/dx^2)]^(dx^2)] + [(1 - dx^2)^(1/dx^2)]^((2dx)^2) +
      [(1 - dx^2)^(1/dx^2)]^((3x)^2) + ... + [(1 - dx^2)^(1/dx^2)]^((nx)^2) + ... )

因為[(1 - dx^2)^(1/dx^2)] -> e^(-1)

上式可近似寫成

上式 = (e^( -dx^2) + e^(-(2dx)^2) + e^(-(3dx)^2) + e^(-(ndx)^2) + ...)dx

讓dx趨近於0,

上式 -> Int(0,inf)[e^(-x^2)dx]

令
c =  Int(0,inf)[e^(-x^2)dx]

c^2 = lim(b->inf) Int(0, b)[Int(0,b)[exp(-(x^2+y^2))dy]dx]

令
r^2 =  x^2+y^2
dxdy = rdrda

c^2 = lim(b->inf) Int(0, pi/2)[Int(0,b)[exp(-r^2)rdr]da] =
lim(b->inf) (pi/2)(-1/2)Int(0,b)[exp(-r^2)d(-r^2)] = pi/4

c = sqrt(pi/4)