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How to prove:

$$\lim_{n \rightarrow \infty} \sqrt [n] {n^2 +n} $$

I am incline to believe it is 1 but all I have tried to prove it had failed so far.

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    $\begingroup$ $n^2\le n^2+n\le 2n^2$. Squeeze... $\endgroup$ – David Mitra Dec 26 '14 at 14:57
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$1 \le \sqrt[n]{n^2+n} \le \sqrt[n]{2n^2} = \sqrt[n]{2} \cdot \sqrt[n]{n}^2 \to 1 \cdot 1^2 = 1$ as $n \to \infty$

Therefore by squeeze theorem...

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$$\lim_{n \rightarrow \infty} \sqrt [n]{n^2 +n} = \lim_{n \rightarrow \infty} \exp\left(\frac{\log(n^2 + n)}{n}\right) = \lim_{n \rightarrow \infty} \exp\left(\frac{2n+1}{n^2 +n}\right)=1.$$

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  • $\begingroup$ Can you please give me some hints for how you got $\frac{2n+1}{n^2+n}$ from $\frac{d\frac{\ln(n^2+n)}{n}}{dx}$? Thanks! I tried both product and quotient rules but I couldn't get rid of the $ln(n^2+n)$ $\endgroup$ – user917099 May 12 '19 at 5:08
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    $\begingroup$ I used L’Hopital rule. The derivative of numerator regarding to $n$ is $$\frac{2n+1}{n^2+n}$$ and the derivative of denominator is $1$. $\endgroup$ – Alex Silva May 12 '19 at 6:38
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Take the logarithm, and get $\log(n^2+n)/n$.
Use L'Hopital's rule.

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  • $\begingroup$ Could you share how you got $log(n^2+n)/n$ from $\lim_{n\to\infty} \sqrt[n]{n^2+n}$? Thanks! $\endgroup$ – user917099 May 5 '19 at 8:34
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    $\begingroup$ $\log(n^2+n)/n=\log(\sqrt[n]{n^2+n})$ by the log laws. That is just the first step. Then, L'Hopital's rule says you differentiate both $\log(n^2+n)$ in the numerator and $n$ in the denominator. That is what Alex Silva has done. $\endgroup$ – Empy2 May 5 '19 at 10:52
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HINT: $$\sqrt[n]{n^2+n}=n^{\frac{1}{n}}(1+n)^{\frac{1}{n}}$$ Also $$n^{\frac{1}{n}}\to1.$$

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  • $\begingroup$ Hmm how do you explain the limit of $(1+n)^\frac{1}{n}$? It's not quite obvious I think.. $\endgroup$ – user21820 Dec 27 '14 at 11:37
  • $\begingroup$ @user21820: $n+1\to\infty$ as $n\to\infty.$ $\endgroup$ – Bumblebee Dec 27 '14 at 12:46
  • $\begingroup$ No that is not enough! That kind of reasoning would imply that $(n^n)^\frac{1}{n} \to 1$ as $n \to \infty$ which is absolutely false. $\endgroup$ – user21820 Dec 27 '14 at 14:45
  • $\begingroup$ Ohh. of course you are right. Let $$y^n=(1+n)$$ Then $$\ln y=\dfrac{\ln n+1}{n}\to 0$$ as $n\to\infty.$ I think this would be enough. Thank you. $\endgroup$ – Bumblebee Dec 27 '14 at 14:53
  • $\begingroup$ Yes it is. But then you might as well have done that on the original expression from the start.. $\endgroup$ – user21820 Dec 27 '14 at 15:27

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