Question

lim стремиться к +бесконечности (4n-5/4n+1)^5n

Answer 1

$\displaystyle \lim_{n \to \infty}\left(\dfrac{4n-5}{4n+1}\right)^{5n}=\lim_{n \to \infty}\left(1-\dfrac{6}{4n+1}\right)^{5n\cdot \frac{-6}{4n+1}\cdot \frac{4n+1}{-6}}=\\ \\ \\ =e^{\lim_{n \to \infty}-\frac{30n}{4n+1}}=e^{-30/4}=e^{-15/2}$