lim x→∞ x^2sin^2 1/x 求极限

解法一：原式=lim(x->∞)[sin²(1/x)/(1/x)²]
=lim(x->∞)[sin(1/x)/(1/x)]²
=1 (以x代换1/x,应用重要极限lim(x->0)(sinx/x)=1);
解法二：（罗比达法）
原式=lim(x->∞)[sin²(1/x)/(1/x)²]
=lim(y->0)(sin²y/y²) (令y=1/x)
=lim(y->0)[2sinycosy/(2y)] (0/0型极限,应用罗比达法则)
=lim(y->0)[sin(2y)/(2y)]
=lim(y->0)[2cos(2y)/2] (0/0型极限,应用罗比达法则)
=lim(y->0)[cos(2y)]
=1.

=lim(y→0)(siny)^2/y2
=1

当 x→∞ 时，1/x→0，即为无穷小量
则有
lim x→∞ x^2sin^2 1/x
=lim x→∞ {x^2[(1/x)^2]}
=1

lim x→∞ x^2sin^2 1/x
= lim x→∞ 〔sin^2 1/x〕／ ( 1/x^2 )
= lim x→∞ [2sin (1/x) * cox (1/x)*(-1/x^2)]/(-2/x^3)
= lim x→∞ sin (1/x) / (1/x)
= lim x→∞ cox(1/x)*(-1/x^2) / (-1/x^2)
= 1