Question

Lim n-> бесконечность n³-7
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4n³+4n²+n

Answer 1

$lim_{n to infty} frac{n^3-7}{4n^3+4n^2+n}=^frac{infty}{infty}= lim_{n to infty} frac{n^3(1-frac{7}{n^3})}{n^3(4+frac{4}{n}+frac{1}{n^2})}=frac{1}{4}$