Question

高数求解~

1.lim(x→0+)(sinx/x+xsin1/x)
2.lim(x→∞）（2x+3/2x+1)的 x+1次方
3.lim(x→0)x-sinx/x的2次方d(1-x^2)=-2xdx

Answer 1

1；e；0 。 解：① lim(x→0＋)[sinx/x＋xsin(1/x)] =1＋0 =1 ② lim(x→∞）[(2x＋3)/(2x＋1)]^( x＋1) =lim(x→∞）[1＋2/(2x＋1)]^(x＋1) =lim(x→∞)[1＋1/(x＋0.5)^(x＋0.5) * lim(x→∞) [1＋1/(x＋0.5)]^0.5 =e*1 =e ③ lim(x→0)(x-sinx)/x² =lim(x→0）(1-cosx)/2x =lim(x→0）(sinx/2) = 0

Answer 2

1.lim(x→0+)(sinx/x+xsin1/x)
=1+0
=1

2.lim(x→∞）（2x+3/2x+1)的 x+1次方
=.lim(x→∞）(1+2/(2x+1)^(x+1)
=.lim(x→∞）(1+1/(x+1/2)^(x+1/2) *[1+1/(x+1/2)]^(1/2)
=e*1
=e

3.lim(x→0)x-sinx/x的2次方
=.lim(x→0）(x-sinx)/x^2
=.lim(x→0）(1-cosx)/2x
=.lim(x→0）(sinx/2)
=0

Answer 3

1、结果为1
2、结果为e
3、结果为0

Answer 4

1;e;e的-2次方