利用洛必达法则求下列极限
来源:百度知道 编辑:UC知道 时间:2024/06/17 21:13:20
(1)lim(x→0)[(e^x)-(e^-x)]/x
(2)lim(x→π/2)cosx/(x-π/2)
(3)lim(x→2)(√x+7)-3/x-2
(4)lim(x→0)sin3x/tan2x
(5)lim(x→+∞)(lnx)^2/x
(6)lim(z→+∞)(x^2)(e^-3x)
请教详解(过程),谢谢!
(2)lim(x→π/2)cosx/(x-π/2)
(3)lim(x→2)(√x+7)-3/x-2
(4)lim(x→0)sin3x/tan2x
(5)lim(x→+∞)(lnx)^2/x
(6)lim(z→+∞)(x^2)(e^-3x)
请教详解(过程),谢谢!
(1)lim(x→0)[(e^x)-(e^-x)]/x
=lim(x→0)[(e^x)+(e^-x)]/1=2
(2)lim(x→π/2)cosx/(x-π/2)
=lim(x→π/2)-sinx/1 = -1
(3)lim(x→2)(√x+7)-3/x-2
=lim(x→2)((1/2)√x+7)^(-1/2)/1 = 1/6
(4)lim(x→0)sin3x/tan2x
=lim(x→0)3cos3x/2(sec2x)^2 = 3/2
(5)lim(x→+∞)(lnx)^2/x
=lim(x→+∞)2(lnx)(1/x)/1
=lim(x→+∞)2(lnx)/x
=lim(x→+∞)2/x = 0
(6)lim(z→+∞)(x^2)(e^-3x)
=lim(z→+∞)(x^2)/(e^3x)
=lim(z→+∞)2x/3(e^3x)
=lim(z→+∞)2/9(e^3x)
=0