<s>[INST] 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?
 [/INST] Remember that $$ \left(\frac{n^2+3}{n^2+5 n-4}\right)^{2n}=\exp\left(2n \ln\left(\frac{n^2+3}{n^2+5 n-4}\right)\right).$$ To compute the limit of the inside, use L'Hospital's rule : 
\begin{align*}
\lim_{n \to +\infty} 2n \ln\left(\frac{n^2+3}{n^2+5 n-4}\right)& =\lim_{n\to +\infty}\frac{\ln\left(\frac{n^2+3}{n^2+5 n-4}\right)}{\frac{1}{2n}}\\
& =\lim_{n\to +\infty}-\frac{2n^2(2n(n^2+5n-4)-(n^2+3)(2n+5))}{(n^2+3)(n^2+5n-4)} \\
& = -10
\end{align*}Indeed the numerator and the denominator are both polynomial of degree $4$ so juste compare the coefficients of $n^4$.
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