Ciara Mcdaniel

2022-07-02

Find the limit:
$\underset{n\to \mathrm{\infty }}{lim}{\left(\frac{3{n}^{2}+2n+1}{3{n}^{2}-5}\right)}^{\frac{{n}^{2}+2}{2n+1}}$

Do you have a similar question?

Charlee Gentry

Expert

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

Still Have Questions?

babyagelesszj

Expert

The standard trick for dealing with ${1}^{\mathrm{\infty }}$ forms is to take logs; it’s very useful if you don’t see anything slicker. Let
$L=\underset{n\to \mathrm{\infty }}{lim}{\left(\frac{3{n}^{2}+2n+1}{3{n}^{2}-5}\right)}^{\frac{{n}^{2}+2}{2n+1}}\phantom{\rule{thickmathspace}{0ex}};$
then
$\begin{array}{rl}\mathrm{ln}L& =\mathrm{ln}\underset{n\to \mathrm{\infty }}{lim}{\left(\frac{3{n}^{2}+2n+1}{3{n}^{2}-5}\right)}^{\frac{{n}^{2}+2}{2n+1}}\\ & =\underset{n\to \mathrm{\infty }}{lim}\mathrm{ln}{\left(\frac{3{n}^{2}+2n+1}{3{n}^{2}-5}\right)}^{\frac{{n}^{2}+2}{2n+1}}\\ & =\underset{n\to \mathrm{\infty }}{lim}\left(\frac{{n}^{2}+2}{2n+1}\right)\mathrm{ln}\left(\frac{3{n}^{2}+2n+1}{3{n}^{2}-5}\right)\phantom{\rule{thickmathspace}{0ex}},\end{array}$
where the second step uses the continuity of the log function. This is an $\mathrm{\infty }\cdot 0$ form, which you can easily convert to a $\frac{0}{0}$ form:
$\underset{n\to \mathrm{\infty }}{lim}\frac{\mathrm{ln}\left(\frac{3{n}^{2}+2n+1}{3{n}^{2}-5}\right)}{\frac{2n+1}{{n}^{2}+2}}\phantom{\rule{thickmathspace}{0ex}}.$
Once you know $\mathrm{ln}L$, recovering L is trivial; just remember to do it!

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