lim(x→1) sin^2(x-1)/x-1 求极限
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lim(x→1) sin^2(x-1)/x-1 求极限
lim(x→1) sin^2(x-1)/x-1 求极限
lim(x→1) sin^2(x-1)/x-1 求极限
洛必塔法则.
原式=lim(x→1) sin(2x-2)/1=0
lim(x->1) [sin(x-1)]^2/(x-1) (0/0)
=lim(x->1) 2sin(x-1).cos(x-1)
=0
o/o型,洛必达法则
原式=lim(x→1) [2sin(x-1)cos(x-1)]/1
=lim(x→1) sin(2x-2)
=sin0
=0
望采纳
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