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→∞)](/uploads/image/z/6859438-70-8.jpg?t=lim+%5B%28x%5E2%2B1%29%2F%28x%5E3%2B1%29%5D+%2A%283%2Bcosx%29+%28n%E2%86%92%E2%88%9E%29)
因为
3+cosx是个有界函数x→∞lim[(x^2+1)/(x^3+1)]=02=
3+cosx是个有界函数x→∞lim[(x^2+1)/(x^3+1)]=02=
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