作业帮求解几个简单的高数习题,请详解.1.lim 0〔sin2x/sin5x] 2.limx～0[tan3x/sin2x]3.limx～0[xcot2x]4.limx～0 [1-cos2x]/[xsinx]5.limx～无穷大 xsin(1/x)
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![求解几个简单的高数习题,请详解.1.lim 0〔sin2x/sin5x] 2.limx～0[tan3x/sin2x]3.limx～0[xcot2x]4.limx～0 [1-cos2x]/[xsinx]5.limx～无穷大 xsin(1/x)](/uploads/image/z/4050980-44-0.jpg?t=%E6%B1%82%E8%A7%A3%E5%87%A0%E4%B8%AA%E7%AE%80%E5%8D%95%E7%9A%84%E9%AB%98%E6%95%B0%E4%B9%A0%E9%A2%98%2C%E8%AF%B7%E8%AF%A6%E8%A7%A3.1.lim+0%E3%80%94sin2x%2Fsin5x%5D+2.limx%EF%BD%9E0%5Btan3x%2Fsin2x%5D3.limx%EF%BD%9E0%5Bxcot2x%5D4.limx%EF%BD%9E0+%5B1-cos2x%5D%2F%5Bxsinx%5D5.limx%EF%BD%9E%E6%97%A0%E7%A9%B7%E5%A4%A7+xsin%281%2Fx%29)
求解几个简单的高数习题,请详解.1.lim 0〔sin2x/sin5x] 2.limx～0[tan3x/sin2x]3.limx～0[xcot2x]4.limx～0 [1-cos2x]/[xsinx]5.limx～无穷大 xsin(1/x)
求解几个简单的高数习题,请详解.
1.lim 0〔sin2x/sin5x]
2.limx～0[tan3x/sin2x]
3.limx～0[xcot2x]
4.limx～0 [1-cos2x]/[xsinx]
5.limx～无穷大 xsin(1/x)
求解几个简单的高数习题,请详解.1.lim 0〔sin2x/sin5x] 2.limx～0[tan3x/sin2x]3.limx～0[xcot2x]4.limx～0 [1-cos2x]/[xsinx]5.limx～无穷大 xsin(1/x)
1.原式=lim(x->0){(2/5)[sin(2x)/(2x)][(5x)/sin(5x)]}
=(2/5)*lim(x->0)[sin(2x)/(2x)]*lim(x->0)[(5x)/sin(5x)]
=(2/5)*1*1 (应用重要极限lim(x->0)(sinx/x)=1)
=2/5;
2.原式=(3/2)lim(x->0){[sin(3x)/(3x)][(2x)/sin(2x)][1/cos(3x)]}
=(3/2)*lim(x->0)[sin(3x)/(3x)]*lim(x->0)[(2x)/sin(2x)]*lim(x->0)[1/cos(3x)]
=(3/2)*1*1*1 (应用重要极限lim(x->0)(sinx/x)=1)
=3/2;
3.原式=lim(x->0)[(x/sinx)*cosx]
=lim(x->0)(x/sinx)*lim(x->0)(cosx)
=1*1 (应用重要极限lim(x->0)(sinx/x)=1);
4.原式=lim(x->0){[(1-cos(2x))/x²]*(x/sinx)}
=2*lim(x->0)[(sinx/x)²*(x/sinx)]
=2*lim(x->0)[(sinx/x)²]*lim(x->0)(x/sinx)
=2*1²*1 (应用重要极限lim(x->0)(sinx/x)=1)
=2;
5.原式=lim(x->0)[sin(1/x)/(1/x)]
=1 (应用重要极限lim(x->0)(sinx/x)=1).
注：以上各题还可以应用罗比达法则.但应用重要极限简洁明了.