Limit of a Sequence involving 3rd root

Question

I'm not finding any way to simplify and solve the following limit: $$ \lim_{n \to \infty} \sqrt{n^2+n+1}-\sqrt[3]{n^3+n^2+n+1} $$

I've tried multiplying by the conjugate, but this give a more complex limit.

Answer 1

$$\begin{align*}&\sqrt{n^2+n+1}-\sqrt[3]{n^3+n^2+n+1}=\\\\ &=\sqrt{n^2+n+1}-n +\left(n-\sqrt[3]{n^3+n^2+n+1}\right)=\\ \\ &=\dfrac{n^2+n+1-n^2}{\sqrt{n^2+n+1}+n } +\dfrac{n^3-n^3-n^2-n-1}{n^2+n(n^3+n^2+n+1)^{\frac{1}{3}}+(n^3+n^2+n+1)^{\frac{2}{3}}}= \\\\ &=\dfrac{n+1}{\sqrt{n^2+n+1}+n } +\dfrac{-(n^2+n+1)}{n^2+n(n^3+n^2+n+1)^{\frac{1}{3}}+(n^3+n^2+n+1)^{\frac{2}{3}}} \underset{n\to\infty}\to \dfrac{1}{2}-\dfrac{1}{3}=\dfrac{1}{6} \end{align*}$$

Answer 2

Notice the $n^{th}$ term of the sequence can be rewritten as $$ \sqrt{n^2+n+1} - \sqrt[3]{n^3 + n^2 + n + 1} = \sqrt{\frac{n^3-1}{n-1}}-\sqrt[3]{\frac{n^4-1}{n-1}} = n \left[\sqrt{\frac{1-n^{-3}}{1-n^{-1}}} - \sqrt[3]{\frac{1-n^{-4}}{1-n^{-1}}}\right] $$ When one expand what's in the square/cubic roots in the rightmost expression as a series in $n^{-1}$, the numerator contributes nothing up to order $O(n^{-3})$. For the denominators, we have

$$\frac{1}{\sqrt{1-n^{-1}}} \sim 1 + \frac12 n^{-1} + O(n^{-2}) \quad\text{ AND }\quad \frac{1}{\sqrt[3]{1-n^{-1}}} \sim 1 + \frac13 n^{-1} + O(n^{-2})$$

As a result, the $n^{th}$ term has the dependence:

$$n \left[ \left(1 + \frac12 n^{-1} + O(n^{-2})\right) - \left(1 + \frac13 n^{-1} + O(n^{-2})\right) \right] = \frac{1}{6} + O(n^{-1}) $$ This immediately tells us the limit is $\frac16$ as $n \to \infty$.

Answer 3

Let $n=\frac{1}{t}$. We will find $$\lim_{t\to 0^+}\frac{\sqrt{1+t+t^2}-\sqrt[3]{1+t+t^2+t^3}}{t}$$ using L'Hospital's Rule. So we need to find $$\lim_{t\to 0^+} \left(\frac{1+2t}{2(1+t+t^2)^{1/2}}-\frac{1+2t+3t^2}{3(1+t+t^2+t^3)^{2/3}}\right).$$ This is $\frac{1}{2}-\frac{1}{3}$.