Question

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

Answer 1

`\displaystyle\lim_(n\to\infty)\left((5n^2+7n)/(4n^2-4)\right)^n=\lim_(n\to\infty)\exp\left(n\ln(5n^2+7n)/(4n^2-4)\right)=\exp\left(\lim_(n\to\infty)n\ln(5n^2+7n)/(4n^2-4)\right)`


`(5n^2+7n)/(4n^2-4)=\frac54+\frac3{2n+2}+\frac1{4n-4}`

As
`n\to\infty`, the above expression approaches
`\frac54`. Meanwhile
`n\ln(5n^2+7n)/(4n^2-4)\to\infty*\ln\left(\frac54\right)=\infty`.

So,


`\displaystyle\lim_(n\to\infty)\left((5n^2+7n)/(4n^2-4)\right)^n=\infty`