lim [(x^2+1)/(x^3+1)] *(3+cosx) (n→∞)
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lim[(x^2+1)/(x^3+1)]*(3+cosx)(n→∞)lim[(x^2+1)/(x^3+1)]*(3+cosx)(n→∞)lim[(x^2+1)/(x^3+1)]*(3+cosx)(n→
lim [(x^2+1)/(x^3+1)] *(3+cosx) (n→∞)
lim [(x^2+1)/(x^3+1)] *(3+cosx) (n→∞)
lim [(x^2+1)/(x^3+1)] *(3+cosx) (n→∞)
因为
3+cosx是个有界函数x→∞lim[(x^2+1)/(x^3+1)]=02=<3+cosx<=4所以原极限=0
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