求极限x趋近于无穷,lim(x^2*arcsin1/x)/4x-1要求用罗比达法则解、、亲们~

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求极限x趋近于无穷,lim(x^2*arcsin1/x)/4x-1要求用罗比达法则解、、亲们~
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求极限x趋近于无穷,lim(x^2*arcsin1/x)/4x-1要求用罗比达法则解、、亲们~
求极限x趋近于无穷,lim(x^2*arcsin1/x)/4x-1
要求用罗比达法则解、、亲们~

求极限x趋近于无穷,lim(x^2*arcsin1/x)/4x-1要求用罗比达法则解、、亲们~
lim(x^2*arcsin1/x)/4x-1,x→∞,1/x→0,等价无穷小代换,1/x→0,1/x~sin(1/x),arcsin1/x→1/x,则lim(x^2*arcsin1/x)/4x-1=limx/4x-1=1/4.

先拆成两项
lim(x^2*arcsin1/x)/4x-1
=lim(x/(4x-1) * lim xarcsin(1/x)
lim(x/(4x-1)=lim 1/4 = 1/4 (L'Hospital)
令t=1/x
则lim xarcsin(1/x)
=lim (t->0)arcsint/t
用L'Hospital
=lim 1/√(1-x^2) / 1
=1
所以原式=(1/4) * 1 = 1/4