A. Granica

$\underset{n\to \infty }{\lim }\frac{n^2-10}{1-5n^2}=\underset{n\to \infty }{\lim }\frac{1-\frac{10}{n^2}}{\frac{1}{n^2}-5}=-\frac{1}{5}$  

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B. Granica

$\underset{n\to \infty }{\lim }\frac{\left(2n-3\right)\left(1-7n\right)}{\left(3+n\right)\left(n+2\right)}=\underset{n\to \infty }{\lim }\frac{\left(2-\frac{3}{n}\right)\left(\frac{1}{n}-7\right)}{\left(\frac{3}{n}+1\right)\left(1+\frac{2}{n}\right)}=\frac{2\cdot \left(-7\right)}{1\cdot 1}=-14$  

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C. Granica

$\underset{n\to \infty }{\lim }\frac{5n^2-1}{2n^3+n+2}=\underset{n\to \infty }{\lim }\frac{5-\frac{1}{n^2}}{2n+\frac{1}{n}+\frac{2}{n^2}}=\frac{5}{\infty }=0$  

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D. Granica

$\underset{n\to \infty }{\lim }\frac{{\left(3n^2-1\right)}^2}{\left(n^3+5\right)\left(1+3n\right)}=\underset{n\to \infty }{\lim }\frac{{\left(3-\frac{1}{n^2}\right)}^2}{\left(1+\frac{5}{n^3}\right)\left(\frac{1}{n}+3\right)}=\frac{{\left(3-0\right)}^2}{1\cdot 3}=\frac{9}{3}=3$  

Liczba 3 jest liczbą pierwszą.

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Prawidłowa odpowiedź to D.