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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?

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up vote 1 down vote accepted

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.

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Limit going to what? If $n \to \infty$, a number less than $\frac12$ is raised to a high power so it goes rapidly to $0$.

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Yes? $(1-\frac{1}{n})^n$ :-) – Listing Dec 21 '11 at 0:30
@Listing: true, but here it doesn't go to $1^\infty$. I'll update. – Ross Millikan Dec 21 '11 at 1:01




$$\lim_{n \to+\infty}n\ln{\frac{2n+3}{n^2}}$$

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So... e raised to that power?? – The Chaz 2.0 Dec 21 '11 at 2:19

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 $n > 5$.

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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 $\displaystyle\lim _{n\rightarrow\infty} y=0.$

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