Mathematics: Lim n- infinite (5n^2+7n/4n^2-4)^n

Lim n- infinite (5n^2+7n/4n^2-4)^n - id5274243012280, bestdiva2

Ответ:

neash19

05.01.2022

$\displaystyle\lim_{n\to\infty}\left(\frac{5n^2+7n}{4n^2-4}\right)^n=\lim_{n\to\infty}\exp\left(n\ln\frac{5n^2+7n}{4n^2-4}\right)=\exp\left(\lim_{n\to\infty}n\ln\frac{5n^2+7n}{4n^2-4}\right)$

$\dfrac{5n^2+7n}{4n^2-4}=\dfrac54+\dfrac3{2n+2}+\dfrac1{4n-4}$

As $n\to\infty$ , the above expression approaches $\dfrac54$ . Meanwhile $n\ln\dfrac{5n^2+7n}{4n^2-4}\to\infty\times\ln\left(\dfrac54\right)=\infty$ .

So,

$\displaystyle\lim_{n\to\infty}\left(\frac{5n^2+7n}{4n^2-4}\right)^n=\infty$

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Ответ:

dedgefield

29.06.2022

Mathematics

2.4 gal of gas per hr

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Lim n-> infinite (5n^2+7n/4n^2-4)^n

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