Question

lim x→∞ x^2sin^2 1/x 求极限

Answer 1

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.

Answer 2

xsin(1/x^2)-x/sin2x
1/x^2→∞,所以sin(1/x^2)在-1到1之间震荡
而x→0，所以xsin(1/x^2)极限是0
x/sin2x=(1/2)*(2x)/sin2x
x→0则2x→0
所以2x/sin2x极限是1
所以原式极限=0-1/2=-1/2

Answer 3

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