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$\underset{n\to \infty }{\lim }\frac{\left(\genfrac{}{}{0ex}{}{n+2}{n-1}\right)}{\frac{1}{2}{n}^3-4n+7}$

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Odpowiedź:

$\underset{n\to \infty }{\lim }\frac{\left(\genfrac{}{}{0ex}{}{n+2}{n-1}\right)}{\frac{1}{2}{n}^3-4n+7}=\frac{1}{3}$

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Wyjaśnienie:

Obliczmy podaną granicę:

$\underset{n\to \infty }{\lim }\frac{\left(\genfrac{}{}{0ex}{}{n+2}{n-1}\right)}{\frac{1}{2}{n}^3-4n+7}=\underset{n\to \infty }{\lim }\frac{\frac{\left(n+2\right)!}{\left(n-1\right)!\left(\left(n+2\right)-\left(n-1\right)\right)!}}{\frac{1}{2}{n}^3-4n+7}=\underset{n\to \infty }{\lim }\frac{\frac{\left(n+2\right)!}{\left(n-1\right)!\left(n+2-n+1\right)!}}{\frac{1}{2}{n}^3-4n+7}=\underset{n\to \infty }{\lim }\frac{\frac{\left(n+2\right)!}{\left(n-1\right)!\cdot 3!}}{\frac{1}{2}{n}^3-4n+7}=\underset{n\to \infty }{\lim }\frac{\frac{{\cancel{\left(n-1\right)!}}^1\cdot n\cdot \left(n+1\right)\cdot \left(n+2\right)}{{\cancel{\left(n-1\right)!}}^1\cdot 6}}{\frac{1}{2}{n}^3-4n+7}=\underset{n\to \infty }{\lim }\frac{\frac{n\cdot \left(n+1\right)\cdot \left(n+2\right)}{6}}{\frac{1}{2}{n}^3-4n+7}=\underset{n\to \infty }{\lim }\frac{n\cdot \left(n+1\right)\left(n+2\right)}{6\cdot \left(\frac{1}{2}{n}^3-4n+7\right)}=\underset{n\to \infty }{\lim }\frac{\left(n^2+n\right)\left(n+2\right)}{3n^3-24n+42}=\underset{n\to \infty }{\lim }\frac{n^3+2n^2+n^2+2n}{3n^3-24n+42}=\underset{n\to \infty }{\lim }\frac{n^3+3n^2+2n}{3n^3-24n+42}=\underset{n\to \infty }{\lim }\frac{\frac{n^3}{n^3}+\frac{3n^2}{n^3}+\frac{2n}{n^3}}{\frac{3n^3}{n^3}-\frac{24n}{n^3}+\frac{42}{n^3}}=\frac{1+0+0}{3-0+0}=\frac{1}{3}$