[tex]\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)[/tex]
[tex]\dfrac{5n^2+7n}{4n^2-4}=\dfrac54+\dfrac3{2n+2}+\dfrac1{4n-4}[/tex]
As [tex]n\to\infty[/tex], the above expression approaches [tex]\dfrac54[/tex]. Meanwhile [tex]n\ln\dfrac{5n^2+7n}{4n^2-4}\to\infty\times\ln\left(\dfrac54\right)=\infty[/tex].
So,
[tex]\displaystyle\lim_{n\to\infty}\left(\frac{5n^2+7n}{4n^2-4}\right)^n=\infty[/tex]
