# A limit question with n-th root

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.

• $n^2\le n^2+n\le 2n^2$. Squeeze... – David Mitra Dec 26 '14 at 14:57

## 4 Answers

$1 \le \sqrt[n]{n^2+n} \le \sqrt[n]{2n^2} = \sqrt[n]{2} \cdot \sqrt[n]{n}^2 \to 1 \cdot 1^2 = 1$ as $n \to \infty$

Therefore by squeeze theorem...

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

• 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 $ln(n^2+n)$ – user917099 May 12 '19 at 5:08
• 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 $1$. – Alex Silva May 12 '19 at 6:38

Take the logarithm, and get $\log(n^2+n)/n$.
Use L'Hopital's rule.

• Could you share how you got $log(n^2+n)/n$ from $\lim_{n\to\infty} \sqrt[n]{n^2+n}$? Thanks! – user917099 May 5 '19 at 8:34
• $\log(n^2+n)/n=\log(\sqrt[n]{n^2+n})$ 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. – Empy2 May 5 '19 at 10:52

HINT: $$\sqrt[n]{n^2+n}=n^{\frac{1}{n}}(1+n)^{\frac{1}{n}}$$ Also $$n^{\frac{1}{n}}\to1.$$

• Hmm how do you explain the limit of $(1+n)^\frac{1}{n}$? It's not quite obvious I think.. – user21820 Dec 27 '14 at 11:37
• @user21820: $n+1\to\infty$ as $n\to\infty.$ – Bumblebee Dec 27 '14 at 12:46
• 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. – user21820 Dec 27 '14 at 14:45
• 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. – Bumblebee Dec 27 '14 at 14:53
• Yes it is. But then you might as well have done that on the original expression from the start.. – user21820 Dec 27 '14 at 15:27