Find the limit of the sequence with $n$th term $x_n = \left(\frac{2n+3}{n^2}\right)^n$ How to find the limit of the sequence with $n^{th}$ term $$x_n = \left(\frac{2n+3}{n^2}\right)^n$$
Is it possible to apply l'Hôpital's rule in this case?
 A: Limit going to what?  If $n \to \infty$, a number less than $\frac12$ is raised to a high power so it goes rapidly to $0$.
A: As usual when the variable appears in the exponent, write it using the rule $a^b = \exp(b \ln a)$ :
$$\exp\left( n \ln\left({2n + 3 \over n^2}\right)\right)$$
and
$$n \ln\left({2n + 3 \over n^2}\right) = n\ln(2n+3) - 2n\ln(n) = n\ln(n)\left( \ln\left(2+{3\over n}\right)-2 \right). $$
It is now easy to conclude: $n \ln(n) \mathop\longrightarrow\limits_{n\rightarrow\infty} \infty$ and $ \ln\left(2+{3\over n}\right)-2\mathop\longrightarrow\limits_{n\rightarrow\infty} \ln 2 -2 < 0$, so $n \ln\left({2n + 3 \over n^2}\right) \mathop\longrightarrow\limits_{n\rightarrow\infty} -\infty$ and $ \exp\left( n \ln\left({2n + 3 \over n^2}\right)\right) \mathop\longrightarrow\limits_{n\rightarrow\infty} 0$.
To have a chance to apply l’Hospital rule, try to put $x = 1/n$ and study the limit in 0, but I think this is not very useful here, as the above method is rather straightforward.
A: well:
$$\left(\frac{2n+3}{n^2}\right)^n=e^{n\ln{\frac{2n+3}{n^2}}}$$
so:
$$\lim_{n \to+\infty}n\ln{\frac{2n+3}{n^2}}$$
A: Find a sequence $y_n$ such that $x_n \leq y_n$ and $\lim\limits_{n \to \infty} y_n = 0$.  Since $0 \leq x_n \forall n$, then we can write $0 \leq \lim\limits_{n \to \infty} x_n \leq \lim\limits_{n \to \infty} y_n = 0$ and it follows that $\lim\limits_{n \to \infty} x_n = 0$.  One such sequence is $5/n$ for $n > 5$.    
A: Let $y=\big(\frac{2n+3}{n^2}\big)^n$, so $\ln y= n\ln \big(\frac{2n+3}{n^2}\big)$. Since $\displaystyle\lim _{n\rightarrow\infty}n\ln(\frac{2n+3}{n^2})=-\infty$, we have $\displaystyle\lim _{n\rightarrow\infty} y=0.$
