lim(x²+1/3x-2)sinπ/2 (x趋于正无穷)=?lim(sinx/2x-xsinπ/x) (x趋于正无穷)=?

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lim(x²+1/3x-2)sinπ/2 (x趋于正无穷)=?lim(sinx/2x-xsinπ/x) (x趋于正无穷)=?
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lim(x²+1/3x-2)sinπ/2 (x趋于正无穷)=?lim(sinx/2x-xsinπ/x) (x趋于正无穷)=?
lim(x²+1/3x-2)sinπ/2 (x趋于正无穷)=?lim(sinx/2x-xsinπ/x) (x趋于正无穷)=?

lim(x²+1/3x-2)sinπ/2 (x趋于正无穷)=?lim(sinx/2x-xsinπ/x) (x趋于正无穷)=?
第一题直接可以看出极限为正无穷
第二题运用无穷比无穷型极限求法:lim(sinx/2x-xsinπ/x)=limsinx/2x+limxsinπ/x
limsinx/2x中|sinx(x->无穷大)|≤1有界,而2x趋于无穷大,所以极限为0
limxsinπ/x=sinπ=0
所以lim(sinx/2x-xsinπ/x)=0
是不是少了几个括号啊?这么简单的?

第二问:lim(sinx/2x)=0,lim(xsinπ/x)=lim(sinπ/x)/(1/x)
=lim(sinπ/x)/(π/x)*π
而sinπ/x与π/x为x趋于正无穷时的等价无穷小,所以原式=0-π=-π