Question

当X趋向于0时求(sinx-tanx)/[(根号下1+X^2)-1][(根号下1+sinX)-1]的极限

Answer 1

Limit [(Sin[x] - Tan[x])/(Sqrt[1 + x^2] - 1) (Sqrt[1 + Sin[x]^2] - 1), x -> 0]
等量无穷小.
=Limit [(Sin[x] - Tan[x])/(Sqrt[1 + x^2] - 1)^2, x -> 0]
应用洛必达法则
= Limit [(Cos[x] - Sec[x]^2)/((2 x (-1 + Sqrt[1 + x^2]))/Sqrt[1 + x^2]), x -> 0]
化简
= Limit [((Cos[x]^3 - 1)/Cos[x]^2)/(2 x (-1 + Sqrt[1 + x^2])), x -> 0]
化简
= Limit [(Cos[x]^3 - 1)/(2 x (-1 + Sqrt[1 + x^2])), x -> 0]
应用洛必达法则
= Limit [(-3 Cos[x]^2 Sin[x])/((2 (1 + 2 x^2 - Sqrt[1 + x^2]))/Sqrt[1 + x^2]), x -> 0] - (2 (-1 - 2 x^2 + Sqrt[1 + x^2]))/Sqrt[1 + x^2]
化简
= Limit [(-3 Sin[x])/(2 (1 + 2 x^2 - Sqrt[1 + x^2])), x -> 0]
应用洛必达法则
= Limit [(-3 Cos[x])/((2 x (-1 + 4 Sqrt[1 + x^2]))/Sqrt[1 + x^2]), x -> 0]
化简
= Limit [-3 /(2 x (-1 + 4 )), x -> 0]
化简
= Limit [-3 /( 6 x), x -> 0]
化简
= Limit [-1 /0, x -> 0]
化简
= -∞
Limit [(Sin[x] - Tan[x])/(Sqrt[1 + x^2] - 1) (Sqrt[1 + Sin[x]^2] - 1), x -> 0]
= Limit [(Sin[x] - Tan[x])/(Sqrt[1 + x^2] - 1) (Sqrt[1 + x^2] - 1), x -> 0]
= Limit [Sin[x] - Tan[x], x -> 0]
= Limit [(Cos[x]Sin[x] - Sin[x])/Cos[x], x -> 0]
= Limit [((Cos[x] – 1) Sin[x])/Cos[x], x -> 0]
= Limit [(0×0/1, x -> 0]
= 0