Question

Вычислите предел последоваьельности lim(((2+3*n+3*n^2)/(3*n-1))+((5-4*n-4*n^2)/(4*n+7)))

Answer 1

$\lim_{n \to \infty} (\frac{2+3n+3n^2}{3n-1}-\frac{5-4n-4n^2}{4n+7}) = \lim_{n \to \infty} (\frac{(2+3n+3n^2)*(4n+7)-(5-4n-4n^2)*(3n-1)}{(3n-1)*(4n+7)}) = \lim_{n \to \infty} (\frac{25n^2+48n+9}{12n^2+17n-7}) = \lim_{n \to \infty} (\frac{25+48/n+9/n^2}{12+17/n-7/n^2}) =\frac{25}{12}$