Calculating exponential limit I've been breaking my mind over this one.
Find the limit.
$\lim\limits_{n \to \infty} (\frac{n^2+3}{n^2+5 n-4})^{2n} $
I know it equals $\frac{1}{e^{10}} $ but can't figure out how to find it.
Help?

Divide your fraction by $n^2$ and look at the asymptotics for $n\rightarrow \infty$
$$\frac{n^2+3}{n^2+5n-4}=\frac{1+\frac{3}{n^2}}{1+\frac{5}{n}-\frac{4}{n^2}}
= 1 - \frac{5}{n} + O(n^{-2}) \sim 1 - \frac{5}{n}$$
And now for the power:
$$\left(\frac{n^2+3}{n^2+5n-4}\right)^{2n}
\sim \left(1 - \frac{5}{n}\right)^{2n}
\sim \left(1 - \frac{10}{2n}\right)^{2n} \rightarrow e^{-10} $$
Another way is to use logarithms
$$2n \ln \frac{n^2+3}{n^2+5n-4}
=2n \ln \left(1 + \frac{7-5n}{n^2+5n-4}\right)$$
$$\sim 2n \frac{7-5n}{n^2+5n-4} = \frac{14n-10n^2}{n^2+5n-4} \rightarrow -10
$$
