利用洛必达求极限,第四题

2025-05-15 22:03:50
推荐回答(3个)
回答1:

你不是算出来了吗😂?极限就是零

回答2:

(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

回答3:

=lim(1/x)/(2x)=lim1/(2x^2)=0