# More and more limits for sequences

So here goes a bit of homework:

$$\lim_{n\to\infty}{\left(\frac{3n^2+2n+1}{3n^2-5}\right)^{\frac{n^2+2}{2n+1}}}$$

Well, this would trivially lead to:

$$\lim_{n\to\infty}{\left(\frac{3+\frac{2}{n}+\frac{1}{n^2}}{3+\frac{5}{n^2}}\right)^{\frac{n\left(1+\frac{2}{n^2}\right)}{2+\frac{1}{n}}}}$$

Which is clearly an indetermination of type "$1^\infty$". Now, I can't really get through this step... Any hints?

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Are you aware that $\lim_{n\to\infty} (1 + 1/n)^n = e$? Can you do some manipulation with your expression to make it look a little more like this one? – user22805 Sep 12 '12 at 2:47

Reduce to the limit for the exponential: $$\lim_{n \to \infty} \left( 1 + \frac{2}{3 n}\frac{1 - \frac{2}{n}}{1 + \frac{5}{3n}} \right)^{\frac{n}{2} - \frac{n-4}{4n+2}} = \underbrace{\lim_{n \to \infty} \left( 1+\frac{2}{3n} \right)^{n/2}}_{\exp\left(\frac{1}{3}\right)} \cdot \underbrace{\lim_{n \to \infty} \left(1 + \frac{2}{3 n}\frac{1 - \frac{2}{n}}{1 + \frac{5}{3n}}\right)^{-\frac{n-4}{4n+2}} }_{1} \cdot \lim_{n \to \infty} \left(\frac{1 + \frac{2}{3 n}\frac{1 + \frac{2}{n}}{1 + \frac{5}{3n}}}{1-\frac{2}{3n}}\right)^{n/2}$$ Using $$\frac{1 + \frac{2}{3 n}\frac{1 + \frac{2}{n}}{1 + \frac{5}{3n}}}{1-\frac{2}{3n}} = 1 - \frac{22}{(3n+2)(3n+5)}$$ we conclude that the last limit also equals to 1.

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I just tried this limit on wolfram alpha and the answer there is $e^{1/3}$. – Integral Sep 12 '12 at 3:02

The standard trick for dealing with $1^\infty$ forms is to take logs; it’s very useful if you don’t see anything slicker. Let

$$L=\lim_{n\to\infty}{\left(\frac{3n^2+2n+1}{3n^2-5}\right)^{\frac{n^2+2}{2n+1}}}\;;$$

then

\begin{align*} \ln L&=\ln\lim_{n\to\infty}{\left(\frac{3n^2+2n+1}{3n^2-5}\right)^{\frac{n^2+2}{2n+1}}}\\ &=\lim_{n\to\infty}\ln{\left(\frac{3n^2+2n+1}{3n^2-5}\right)^{\frac{n^2+2}{2n+1}}}\\ &=\lim_{n\to\infty}\left(\frac{n^2+2}{2n+1}\right)\ln\left(\frac{3n^2+2n+1}{3n^2-5}\right)\;, \end{align*}

where the second step uses the continuity of the log function. This is an $\infty\cdot 0$ form, which you can easily convert to a $\frac00$ form:

$$\lim_{n\to\infty}\frac{\ln\left(\frac{3n^2+2n+1}{3n^2-5}\right)}{\frac{2n+1}{n^2+2}}\;.$$

Once you know $\ln L$, recovering $L$ is trivial; just remember to do it!

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$$\frac{3n^2+2n+1}{3n^2-5}=1+\frac{2n+6}{3n^2-5}=1+\frac{2}{\frac{3n^2-5}{n+3}}\Longrightarrow$$

$$\Longrightarrow\left(\frac{3n^2+2n+1}{3n^2-5}\right)^{\frac{n^2+2}{2n+1}}=\left[\left(1+\frac{2}{\frac{3n^2-5}{n+3}}\right)^{\frac{3n^2-5}{n+3}}\right]^\frac{n^3+3n^2+2n+6}{6n^3+3n^2-10n-5}\Longrightarrow$$

Well, now the inner limit must be well-known, and for the exterior one just check the exponent behaves as $\,1/6\,$ for large values of $\,n\,$ . The limit indeed is $\,e^{1/3}\,$

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