Calculating exponential limit

Question

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?

Answer 1

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

$$2n \ln \frac{n^2+3}{n^2+5n-4} =2n \ln \left(1 + \frac{7-5n}{n^2+5n-4}\right)$$ $$\sim 2n \frac{7-5n}{n^2+5n-4} = \frac{14n-10n^2}{n^2+5n-4} \rightarrow -10 $$

Answer 2

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 $n^4$.

Answer 3

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 $$ \lim_{n\to\infty}e^{f(n)}=e^{-10} $$