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I have two limit problems which are quite similar, so I've put them both into this post.

  1. $((4^{10}+2^{n})^{\frac{1}{n}})$
  2. $((3n^{2}+n)^{\frac{1}{n}})$

Attempt:

For 1, I'm not sure if this is permissible. I know that $\lim_{n \to \infty}(x^{n}+y^{n})^{\frac{1}{n}} = \max\{x,y\}$. Obviously $4^{10}$ is a constant so it's not quite in the same form, but I believe it is still okay to use this result, and so the limit is $2$.

For 2, I've got $((3n^{2}+n)^{\frac{1}{n}}) = (n(3n+1)^{\frac{1}{n}})$, but then I'm not too sure how to proceed.

I also thought of taking the log and proceeding that way, however in the notes I'm learning from they haven't covered l'hopital yet, so I'm trying to not use it.

Thanks.

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  • $\begingroup$ you mean $((3n^{2}+n)^{\frac{1}{n}}) = (n(3n+1))^{\frac{1}{n}}$ ? $\endgroup$ – J. W. Tanner Nov 4 at 15:48
  • $\begingroup$ Yes typo, thanks. $\endgroup$ – AnalysisLearner Nov 4 at 15:59
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For the first one we have

$$(4^{10}+2^{n})^{\frac{1}{n}}=2\left(\frac{4^{10}}{2^n}+1\right)^{\frac{1}{n}}\to 2\cdot 1^0 =2$$

and

$$(3n^{2}+n)^{\frac{1}{n}}=e^{\log(3n^{2}+n)^{\frac{1}{n}}}=e^{\frac{\log (3n^2+n) }{n}}\to e^0= 1$$

recall indeed that $\frac{\log n}n \to 0$ and therefore

$$\frac{\log (3n^2+n) }{n}\le \frac{\log (3n^2+6n+3) }{n}=\frac{\log (3(n+1)^2) }{n}=\frac{2\log (n+1)+\log 3 }{n}=$$

$$=2\frac{n+1}n\frac{\log (n+1) }{n+1}+\frac{\log 3}n\to 2 \cdot 1 \cdot 0 + 0 =0$$

As an alternative for the second we can also use that

$$n^\frac1n \le(3n^{2}+n)^{\frac{1}{n}} \le (3n^{2}+n^2)^{\frac{1}{n}}=(4n^{2})^{\frac{1}{n}}$$

and conclude in a similar way by squeeze theorem.

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  • $\begingroup$ For the first one, where do you get the factor of $2$ from in the first equality? For the second one where does the $2$ in $2 \frac{n+1}{n}$ come from? As an aside, how would one prove the result about $ \frac{\log n}{n} \to 0$ (I'll have a go myself using definition of null sequence / convergence). Thanks (again!). $\endgroup$ – AnalysisLearner Nov 4 at 16:14
  • $\begingroup$ @AnalysisLearner It is a simple step $$(4^{10}+2^{n})^{\frac{1}{n}}=\left(2^n \left(\frac{4^{10}}{2^n}+1\right)\right)^{\frac{1}{n}}=(2^n)^\frac1n\left(\frac{4^{10}}{2^n}+1\right)^{\frac{1}{n}}=2\left(\frac{4^{10}}{2^n}+1\right)^{\frac{1}{n}}$$ $\endgroup$ – user Nov 4 at 16:53
  • $\begingroup$ @AnalysisLearner ALso the second one is a simple step $$\frac{2\log (n+1)}{n}=2\frac{n+1}n\frac{\log (n+1) }{n+1}$$ I've noticed now tere is a small typo, I fix that. $\endgroup$ – user Nov 4 at 16:55
  • $\begingroup$ @AnalysisLearner A way to prove that is by $ \frac{\log x}{x} \to 0$ which by $x=e^y \to \infty$ becomes $ \frac{\log e^y}{e^y}=\frac y{e^y} \to 0$ . $\endgroup$ – user Nov 4 at 16:59
  • $\begingroup$ @AnalysisLearner Refer also to the related OP. $\endgroup$ – user Nov 4 at 17:01
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Take $\varepsilon>0$. Then $\lim_{n\to\infty}(2+\varepsilon)^n=\infty$ and therefroe $(2+\varepsilon)^n>4^{10}$, if $n\gg1$. But then, if $n$ is large enough,$$(2^n)^{1/n}<(4^{10}+2^n)^{1/n}<\bigl((2+\varepsilon)^n+2^n\bigr)^{1/n}.$$Since $\lim_{n\to\infty}(2^n)^{1/n}=2$ and$$\lim_{n\to\infty}\bigl((2+\varepsilon)^n+2^n\bigr)^{1/n}=\max\{2+\varepsilon,2\}=2+\varepsilon,$$it is easy now to prove that the limit of your sequence is $2$.

For the other sequence, you can use the fact that $3n^2<3n^2+n<4n^2$ and that$$\lim_{n\to\infty}(3n^2)^{1/n}=\lim_{n\to\infty}(4n^2)^{1/n}=1.$$

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  • $\begingroup$ We can use any constant $>1$ instead of $2+\varepsilon$. Then you don't need the limit trick. $\endgroup$ – Milten Nov 4 at 15:53
  • $\begingroup$ I mean the limit $\varepsilon\to0$. The limit argument for large $n$ is good. $\endgroup$ – Milten Nov 4 at 15:57

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