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.