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?
 A: 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 $n^4$.
A: 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
$$2n \ln \frac{n^2+3}{n^2+5n-4}
=2n \ln \left(1 + \frac{7-5n}{n^2+5n-4}\right)$$
$$\sim 2n \frac{7-5n}{n^2+5n-4} = \frac{14n-10n^2}{n^2+5n-4} \rightarrow -10
$$
A: Consider the function
$$
f(x)=2x\log\frac{x^2+3}{x^2+5x-4}
$$
(where log is the natural logarithm). Then your limit is
$$
\lim_{n\to\infty}e^{f(n)}
$$
so we can as well compute the limit at $\infty$ of $f(x)$. However, it's much better to compute limits at $0$, because we know derivatives; so set $x=1/t$ and transform the function into
$$
g(t)=2\frac{\log\dfrac{1+3t^2}{1+5t-4t^2}}{t}
$$
and notice that 
$$
\lim_{x\to\infty}f(x)=
\lim_{t\to0^+}g(t)
$$
which is just the derivative at $0$ of
$$
G(t)=2\log\dfrac{1+3t^2}{1+5t-4t^2}=2\log(1+3t^2)-2\log(1+5t-4t^2)
$$
Since
$$
G'(t)=2\frac{6t}{1+3t^2}-2\frac{5-8t}{1+5t-4t^2}
$$
we have $G'(0)=-10$. Thus your original limit is
$$
\lim_{n\to\infty}e^{f(n)}=e^{-10}
$$
