=limx→2(x+2)[(x-2)/sin(x-2)],
当x→2时,x-2→0,x+2→4,
故x→2时,有(x-2)/sin(x-2)→1.
原式=4
=lim[(x+2)(x-2)]/sin(x-2)
x-->2
x-2--->0
lim(x-2)/sinx=1
lim[(x+2)(x-2)]/sin(x-2)
=lim(x+2)=4
这里利用了那个重要的极限limsinx/x=1 (x-->0)
其实是以t=x-2作为一个整体的
而x^2是无穷小
所以x趋于0时，x^2sin2/x
是无穷小
故limx^2sin2/x
x→0
的极限=0
limx→2(x^2-4)\/sin(x-2),求极限
limx→2(x^2-4)\/sin(x-2)=limx→2(x+2)[(x-2)\/sin(x-2)],当x→2时,x-2→0,x+2→4,故x→2时,有(x-2)\/sin(x-2)→1.原式=4
limx→2(x^2-4)\/sin(x-2),求极限
limx→2(x^2-4)\/sin(x-2)=limx→2(x+2)[(x-2)\/sin(x-2)],当x→2时,x-2→0,x+2→4,故x→2时,有(x-2)\/sin(x-2)→1.原式=4
求解lim x→2 x^2-4\/sin(x-2),要过程,谢
lim(x→2) (x＾2-4)\/sin(x-2)=lim(x→2) (x+2)[(x-2)\/sin(x-2)]∵ x→2 ,t=x-2→0 ,t\/sint -> 1 (重要极限）= （2+2）* 1 = 4
求极限limx→2 (x²-4)\/sin(x-2)
0\/0型极限，用第一个重要极限公式求。
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lim x->2 sin(x^2-4)\/x-2=过程啊
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求极限lim(x→∞)2x^2-x\/x^2+4
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证明: limx→2(x^2)=4(详细的证明过程)
回答：limx→2(x^2)=lim⊿x→0<(x+⊿x)^2-x^2>\/⊿x=lim⊿x→02X+⊿x=2x=4
lim x→2 x^2-4\/x-2 求函数的极限,要过程,怎么求?
^x=lim[1+4\/(x-2)]^(x-2) x→∞ =lim[1+4\/(x-2)]^(4*(x-2)\/4)=e^4 x→∞ lim（x+2\/x-2)^x=e^lim x(ln(x+2)\/(x-2))=e^lim [(ln(x+2)-ln(x-2)]\/(1\/x) x→∞ =e^lim [1\/(x+2)-1\/(x-2)]\/(-1\/x²) =e^lim {-4\/[(x+2)(x-2...
limx->2 sin(x^2-4)\/x-2=??
limx->0e^x^2-e^(2-2cosx)\/x^4 =lim[2xe^x2+2sinxe^(2-2cosx)]\/4x3 =(1\/2)lim [e^x2+e^(2-2cosx)]\/x2 =(1\/2)lim [e^x2-1+e^(2-2cosx)-1]\/x2 =(1\/2){lim [e^x2-1]\/x2+[e^(2-2cosx)-1]\/x2} =(1\/2){1+lim [2-2cosx]\/x2} =(1\/2)...
