Find $\lim_{n \to \infty } \sqrt[3]{n^3+1} - \sqrt{n^2+1}$

Question

Find $$\lim_{n \to \infty} \sqrt[3]{n^3+1}-\sqrt{n^2+1}$$

I already tried to use the Sqeeze theorem on it, but I just was not able to find some reasonable upper series for it, only lower:

$$n\sqrt[3]{1+\frac{1}{n^3}}-n\sqrt{1+\frac{1}{n ^2}}$$ $$n\left(\sqrt[3]{1+\frac{1}{n^3}}-\sqrt{1+\frac{1}{n ^2}}\right)$$ $$\left(\sqrt[3]{1+\frac{1}{n^3}}-\sqrt{1+\frac{1}{n ^2}}\right) \leq n\left(\sqrt[3]{1+\frac{1}{n^3}}-\sqrt{1+\frac{1}{n ^2}}\right)$$

Is there anyone who can give me a hint as to how to solve it?

The lower bound in question is incorrect, try it for $n=1,2,\dots$... reason is the $\left(\sqrt[3]{1+\frac{1}{n^3}}-\sqrt{1+\frac{1}{n ^2}}\right)$ is negative... — Sil, Oct 21 '17 at 12:10
Put $f(x)=\root{3}\of{1+x^3}-\sqrt{1+x^2}$. Then you want the limit of $(f(1/n)-f(0))/(1/n)$. What is the limit of $(f(x)-f(0))/x$ as $x\to 0$, $x\not =0$ ? — Kelenner, Oct 21 '17 at 12:15
More general, https://math.stackexchange.com/questions/2482868/solving-lim-x-to-infty-left-left-1xp1-right-frac1p1-l/2482926?noredirect=1#comment5129780_2482926 — Guy Fsone, Oct 21 '17 at 15:39
This question might also be helpful: Limits: How to evaluate $\lim\limits_{x\rightarrow \infty}\sqrt[n]{x^{n}+a_{n-1}x^{n-1}+\cdots+a_{0}}-x$ — Martin Sleziak, Feb 08 '22 at 11:24

user236182 · Accepted Answer · 2017-10-21T12:41:37.280

$$\lim_{n \to \infty} (\sqrt[3]{n^3+1}-\sqrt{n^2+1})=$$

$$=\lim_{n\to \infty}((\sqrt[3]{n^3+1}-n)-(\sqrt{n^2+1}-n))=$$

Use the formula/identity $$a^k-b^k=(a-b)(a^{k-1}+a^{k-2}b+\cdots+b^{k-1}), k\ge 2, k\in\mathbb Z, a,b\in\mathbb R$$

with, e.g., $a=\sqrt[3]{n^3+1}$, $b=n$, $k=3$, etc.

$$=\lim_{n\to \infty}\left(\frac{(\sqrt[3]{n^3+1})^3-n^3}{(\sqrt[3]{n^3+1})^2+n\sqrt[3]{n^3+1}+n^2}-\\ -\frac{(\sqrt{n^2+1})^2-n^2}{\sqrt{n^2+1}+n}\right)=$$

$$=\lim_{n\to \infty}\left(\frac{1}{(\sqrt[3]{n^3+1})^2+n\sqrt[3]{n^3+1}+n^2}-\\ -\frac{1}{\sqrt{n^2+1}+n}\right)=0-0=0$$

score 1 · Answer 2 · answered Oct 21 '17 at 12:15

Find $$\lim_{n \to \infty} \sqrt[3]{n^3+1}-\sqrt{n^2+1}$$

$$n\sqrt[3]{1+\frac{1}{n^3}}-n\sqrt{1+\frac{1}{n ^2}}$$ $$n\left(\sqrt[3]{1+\frac{1}{n^3}}-\sqrt{1+\frac{1}{n ^2}}\right)$$ $$n\left(\left({1+\frac{1}{n^3}}\right)^{1/3}-\left({1+\frac{1}{n^2}}\right)^{1/2}\right)$$

As $n \to \infty$,this can be approximated as

$$n\left(\left({1+\frac{1}{3n^3}+other terms}\right)-\left({1+\frac{1}{2n^2}}+other terms\right)\right)$$

$$n\left({\frac{1}{3n^3}-\frac{1}{2n^2}}+other terms\right)$$

$${\frac{1}{3n^2}-\frac{1}{2n}}+other terms$$

As $n \to \infty$, all terms become 0. So answer is 0.

score 1 · Answer 3 · answered Oct 21 '17 at 12:57

Consider $$ f(x)=\sqrt[3]{1+x^3}-\sqrt{1+x^2} $$ Then $$ f'(x)=\frac{3x^2}{3\sqrt[3]{(1+x^3)^2)}}-\frac{2x}{2\sqrt{1+x^2}} $$ and therefore $f'(0)=0$. Therefore $$ 0=f'(0)=\lim_{x\to0^+}\frac{\sqrt[3]{1+x^3}-\sqrt{1+x^2}}{x}= \lim_{t\to\infty}\bigl(\sqrt[3]{t^3+1}-\sqrt{t^2+1}\bigr) $$ with the substitution $t=1/x$.

Find $\lim_{n \to \infty } \sqrt[3]{n^3+1} - \sqrt{n^2+1}$

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