Question

高数极限和定积分问题~急~

x→0时，求limex^2-1-x^2/x^2(ex^2-1)

∫sin3xsin5xdx=?

X→∞时，求lim（2/∏-arctantx)^x
第3个问题应该是lim[(2arctanx)/∏]^x

Answer 1

1.
lim （e^x^2-1-x^2)/x^2(e^x^2-1)
=lim （e^x^2-1-x^2)/x^4 (e^x^2-1等价无穷小为x^2)
=lim 2x(e^x^2-1)/4x^3 (洛必达法则)
=lim x*x^2/2x^3 (e^x^2-1等价无穷小为x^2)
=1/2

2.
sin3xsin5x=[cos(5x-3x)-cos(5x+3x)]/2=[cos2x-cos8x]/2

∫sin3xsin5xdx=1/2∫[cos2x-cos8x]dx
=(1/2)*[sin2x/2-sin8x/8]
=sin2x/4-sin8x/16

3.
0<2/∏-arctantx<1,指数又趋于正无穷

∴lim（2/∏-arctantx)^x =0

Answer 2

罗比塔法则：
上下各自求导
lim(2xex^2-2x)/(2x(ex^2-1)+x^2ex^2)
=lim(2ex^2-2)/(2(ex^2-1)+xex^2)
继续求导：
=lim4xex^2/(4xex^2+2x^2ex^2)
=lim4ex^2/(4ex^2+2xex^2)
=4/(4+0)
=1

积化和差
∫sin3xsin5xdx=
∫1/2(cos2x-cos8x)dx=
1/4sin2x-1/16sin8x+c

lim（2arctanx/π)^x
=lim（1+(2arctanx-π/π))^x
=lim[（1+(2arctanx-π/π))^[1/(2arctanx-π/π))](x(2arctanx-π/π))
=e^(x(2arctanx-π/π))

limx(2arctanx-π/π)
=lim(2arctanx-π/π)/(1/x)
罗比塔法则：
上下各自求导
=lim(2/π(1+x^2))/(-1/x^2)

=lim(-2x^2/π(1+x^2))
=-2/π

所以原式=e^(-2/π)