lim x-->正无穷 (x^2 sin1/x) 怎么算?

lim x-->正无穷 (x^2 sin1/x) 怎么算?
lim x-->正无穷 (x^2 sin1/x) 怎么算?

lim x-->正无穷 (x^2 sin1/x) 怎么算?
lim【x→+∞】[(x^2)sin(1/x)]
=lim【x→+∞】[sin(1/x)]/[1/(x^2)]
这是0/0型的极限,应用洛必达法则,有：
lim【x→+∞】[sin(1/x)]/[1/(x^2)]
=lim【x→+∞】[-1/(x^2)]cos(1/x)/[-2/(x^3)]
=lim【x→+∞】[cos(1/x)]/(2/x)
=lim【x→+∞】[xcos(1/x)]/2
=+∞

令F(x)=1/(x^2)
f(x)=sin1/x
则在x趋向于无穷大时，f(x)、F(x)趋向于0；
且在|x|>N时，f'(x)、F'(x)都存在且F'(x)！=0
即由洛必达法则得lim x-->+∞ (x^2 sin1/x)=lim x-->+∞((-1/(x^2)*cos(1/x))/(-2/(x^3)))=lim x-->+∞(1/2)*x=+∞