sinx*x^2当x趋向于无穷时是去穷大量吗是无界函数吗，判断原因是什么
\u51fd\u6570y=(sinx+1)/(1+x²) \u8fd9\u4e2a\u4e3a\u4ec0\u4e48\u662f\u6709\u754c\u51fd\u6570\uff0c\u600e\u4e48\u5224\u65ad\u7684\u56e0\u4e3a|sinx+1|\u22642 (1)
1+x²\u22651
0<1/|x²+1|\u22641 (2)
(1)*(2)
|sinx+1|/|x²+1|\u22642
\u6240\u4ee5 |(sinx+1)/(1+x²)|\u22642
\u6240\u4ee5 \u51fd\u6570y=(sinx+1)/(1+x²) \u662f\u6709\u754c\u51fd\u6570
\u2235f\uff08x\uff09\uff1dxsinx\uff0c\u3000\u2234f\uff08x\uff09/x\uff1dsinx\u3002
\u663e\u7136\uff0c\uff0d1\u2266sinx\u22661\uff0c\u3000\u2234\uff0d1\u2266f\uff08x\uff09/x\u22661\uff0c\u3000\u53c8x\uff1e0\uff0c\u3000\u2234\uff0dx\u2266f\uff08x\uff09\u2266x\u3002
\u2235x\u7684\u53d6\u503c\u662f\u4e0a\u65e0\u754c\u7684\uff0c\u3000\u2234f\uff08x\uff09\u65e2\u4e0b\u65e0\u754c\uff0c\u4e5f\u4e0a\u65e0\u754c\uff0c\u3000\u2234f\uff08x\uff09\u662f\u65e0\u754c\u51fd\u6570\u3002
取x=2kπ(k为整数)则f(x)=0
根据定义显然易知不是无穷大量
再取x=2kπ+π/2得f(x)=2kπ+π/2
易得为无界函数
当x=n*pi时，上式为零，即当x趋于无穷大时，函数极限不存在，不谓无穷大，再者，当x=n*pi/2时，函数趋于正负无穷大，显然无界
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