0

Find, if it exists, the following limit: $\displaystyle\lim_{n\to\infty}\left(\sqrt{\frac{9^n}{n^2}+\frac{3^n}{5n}+2}-\frac{3^n}{n}\right)$.

Daniel W. Farlow
  • 21,614
  • 25
  • 56
  • 98
Cristopher
  • 550
  • 3
  • 14
  • 1
    What are you doing here? Asking questions and then immediately posting their answers? Not nice, please stop doing this. – Alex M. Jul 06 '15 at 21:07
  • 2
    @AlexM. For your information, it's okay to post questions and their answers so that other people or myself can find it and learn from it. See: https://blog.stackexchange.com/2011/07/its-ok-to-ask-and-answer-your-own-questions/ So consider things more carefully before voting. – Cristopher Jul 06 '15 at 21:10
  • @Cristopher That is nice when one really faces an obstacle then finds the answer himself so he can clear up the train of thought. However, as things are now, it seems like it's writing a solution manual, which is not really good. – Hasan Saad Jul 06 '15 at 21:41
  • 1
    @DanielW.Farlow It seems so... what a shame... this site is to help people learn math, after all. – Cristopher Jul 06 '15 at 21:43
  • @Cristopher For what it's worth, I think questions like the one you just asked are not worth posting Q&A for (in my humble opinion). Why? It seems to be something you could simply feed into W|A and have an answer churned out with what value? One of the main issues, I believe, is the extremely high volume of questions MSE gets on a daily basis, making more or less "not special" questions like this one of less value. Don't know if that makes sense, but I imagine that is how some other users feel. I'd encourage a Q&A when facing something extra challenging or unique. – Daniel W. Farlow Jul 06 '15 at 21:46
  • @DanielW.Farlow As far as I know, W|A just gives you the result, it doesn't tell you _how_ to arrive at it. I only wanted to help... – Cristopher Jul 06 '15 at 21:51
  • @Cristopher W|A will give you full solutions if you have Mathematica 9 or greater or Wolfram Alpha Pro, a complete waste of resources and just a money-grabbing douche move by Wolfram. It's just the *nature* of the problem that I am talking about. For an instance of where Q&A is actually a good thing, I'd recommend reading [this post](http://math.stackexchange.com/questions/1196303/fake-induction-proofs) from a while back. Those are the kind of problems more appropriate for the Q&A style in my opinion. – Daniel W. Farlow Jul 06 '15 at 21:53
  • @DanielW.Farlow I see. I guess that's one way of seeing it. I still think it should be fine. I would like to read some moderator's opinions, though. – Cristopher Jul 06 '15 at 21:57
  • As a side note, I have Mathematica 10, and I just asked it to compute the limit. It says: "step-by-step solution unavailable". – Cristopher Jul 06 '15 at 22:11

2 Answers2

2

There is a more efficient way than the approach posted as an answer by the OP. Here, we simply note that

$$\begin{align} \left(y^2+\frac15 y+2\right)^{1/2}&=y\left(1+\frac{1}{10}y^{-1}+O\left(y^{-2}\right)\right)\\\\ &=y+\frac{1}{10}+O\left(y^{-1}\right) \end{align}$$

so that we have immediately

$$\begin{align} \left(y^2+\frac15 y+2\right)^{1/2}-y&=\frac{1}{10}+O\left(y^{-1}\right)\\\\ &\to \frac{1}{10}\,\,\text{as}\,\,y\to \infty \end{align}$$

Mark Viola
  • 166,174
  • 12
  • 128
  • 227
1

$\displaystyle\lim_{n\to\infty}\left(\sqrt{\frac{9^n}{n^2}+\frac{3^n}{5n}+2}-\frac{3^n}{n}\right)=L$

Let $y=\dfrac{3^n}{n}$. If $n\to\infty$ then $y\to\infty$. Also, note that $\left(\dfrac{3^n}{n}\right)^2=\dfrac{9^n}{n^2}=y^2$. Hence:

\begin{align} L &= \displaystyle\lim_{y\to\infty}\left(\sqrt{y^2+\frac{1}{5}y+2}-y\right) \\ &=\lim_{y\to\infty}\left[\left(\sqrt{y^2+ \frac{1}{5}y+2}-y\right)\cdot\frac{\sqrt{y^2+\frac{1}{5}y+2}+y}{\sqrt{y^2+\frac{1}{5}y+2}+y}\right] \\ &=\lim_{y\to\infty}\frac{y^2+\frac{1}{5}y+2-y^2}{\sqrt{y^2+\frac{1}{5}y+2}+y} \\ &=\lim_{y\to\infty}\frac{\frac{1}{5}y+2}{\sqrt{y^2(1+\frac{1}{5y}+ \frac{2}{y^2})}+y} \\ &=\lim_{y\to\infty}\frac{\frac{1}{5}y+2}{y\sqrt{1+\frac{1}{5y}+\frac{2}{y^2}}+y} \\ &=\lim_{y\to\infty}\frac{y(\frac{1}{5}+\frac{2}{y})}{y\left(\sqrt{1+\frac{1}{5y}+\frac{2}{y^2}}+1\right)} \\ &= \lim_{y\to\infty}\frac{\frac{1}{5}+\frac{2}{y}}{\sqrt{1+\frac{1}{5y}+\frac{2}{y^2}}+1}=\frac{\frac{1}{5}}{2} \\ &=\frac{1}{10} \end{align}

Leucippus
  • 22,756
  • 140
  • 38
  • 83
Cristopher
  • 550
  • 3
  • 14