Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
1  
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
add comment

3 Answers

up vote 1 down vote accepted

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.

share|improve this answer
    
I just tried this limit on wolfram alpha and the answer there is $e^{1/3}$. –  Integral Sep 12 '12 at 3:02
    
Great solution. However, this is clearly tagged homework, which suggests that the OP might not want anything quite so complete. –  user22805 Sep 12 '12 at 3:03
    
@DavidWallace You are right. I should have been more careful. –  Sasha Sep 12 '12 at 3:05
add comment

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!

share|improve this answer
add comment

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

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

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 $\,e^{1/3}\,$

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.