Ciara Mcdaniel

Answered

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?

Recalculate according to your conditions!

Answer & Explanation

Charlee Gentry

Expert

2022-07-03Added 19 answers

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

2022-07-04Added 7 answers

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!

Free Math Solver

Help you to address certain mathematical problems

Ask your question.
Get your answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?