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I need help finding the following limit: $$\lim\limits_{n\to\infty} \sqrt[10]{n^2+2} - \sqrt[10]{n^2+1}$$

I usually know how to solve these kind of tasks for example if $\lim\limits_{n\to\infty} \sqrt[4]{n^2+2} - \sqrt[4]{n^2+1}$. We can just multiplicate with $\frac{ \sqrt[4]{n^2+2} + \sqrt[4]{n^2+1}}{\sqrt[4]{n^2+2} + \sqrt[4]{n^2+1}}$ and so one until the root disappers. But what should I do here?

Is there anyone who could give me an advice? I would be very grateful.

Gary
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Analysis_Mark
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5 Answers5

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Since, when $a,b\in\Bbb R$,$$a^{10}-b^{10}=(a-b)\sum_{k=0}^9a^kb^{9-k},$$you have\begin{align}\sqrt[10]{n^2+2}-\sqrt[10]{n^2+1}&=\frac{(n^2+2)-(n^2+1)}{\sum_{k=0}^9\sqrt[10]{n^2+2}^k\sqrt[10]{n^2+1}^{9-k}}\\&=\frac1{\sum_{k=0}^9\sqrt[10]{n^2+2}^k\sqrt[10]{n^2+1}^{9-k}},\end{align}and therefore your limit is equal to $0$.

José Carlos Santos
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    Why are we suddenly dividing this by the sum? And where did $(n^2+2)-(n^2+1)$ go? – Analysis_Mark Mar 26 '22 at 20:57
  • I am dividing by the sum because that's my approach to find your limit. And I used the fact that $(n^2+2)-(n^2+1)=1$. – José Carlos Santos Mar 26 '22 at 21:23
  • jeah but I can't really say why this division is = $\sqrt[10]{n^2+1} - \sqrt[10]{n^2+1}$ – Analysis_Mark Mar 26 '22 at 21:33
  • In the equality$$a-b=\frac{a^{10}-b^{10}}{\sum_{k=0}^ma^kb^{9-k}},$$take $a=\sqrt[10]{n^2+2}$ and $b=\sqrt[10]{n^2+1}$ and then that equality becomes$$\sqrt[10]{n^2+2}-\sqrt[10]{n^2+1}=\frac{(n^2+2)-(n^2+1)}{\sum_{k=0}^9\sqrt[10]{n^2+2}^k\sqrt[10]{n^2+1}^{9-k}}.$$ – José Carlos Santos Mar 26 '22 at 21:35
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$$ \begin{align} \sqrt[10]{n^2+2}-\sqrt[10]{n^2+1} &=n^{1/5}\left(\left(1+\frac2{n^2}\right)^{1/10}-\left(1+\frac1{n^2}\right)^{1/10}\right)\tag1\\ &\le n^{1/5}\left(\left(1+\frac2{10n^2}\right)-1\right)\tag2\\ &=\frac15n^{-9/5}\tag3 \end{align} $$ Explanation:
$(1)$: factor $n^{1/5}$ out front
$(2)$: apply Bernoulli's Inequality to the left term
$\phantom{\text{(2):}}$ the right term is at least $1$
$(3)$: simplify

Thus, $$ \lim_{n\to\infty}\left(\sqrt[10]{n^2+2}-\sqrt[10]{n^2+1}\right)=0\tag4 $$

Clarification

In a comment:

Wait a minute. Bernoullis inequality is this $(1+x)^n\geq (1+nx)$. But you did it the other way around and said $(1+\frac{1}{n^2})^{\frac{1}{10}}$$\leq 1+\frac{1}{10n^2}$...

For an exponent $a\ge1$, Bernoulli's Inequality is $$ (1+x)^a\ge1+ax\tag5 $$ By substituting $x\mapsto\frac xa$ and raising both sides to the $\frac1a$ power, we get $$ 1+\frac xa\ge(1+x)^{\frac1a}\tag6 $$ which, by setting $b=\frac1a$, says that for $b\in(0,1]$, $$ 1+bx\ge(1+x)^b\tag7 $$ $(7)$, which only looks like the reverse of Bernoulli, is what is used in $(2)$ above.

robjohn
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  • Using Bernoulli's Inequality, we can get $$1+\frac1{10n^2+10}\le\left(1+\frac1{n^2}\right)^{1/10}\le1+\frac1{10n^2}$$ and $$1+\frac1{5n^2+10}\le\left(1+\frac2{n^2}\right)^{1/10}\le1+\frac1{5n^2}$$ This gives not only $$\begin{align}\sqrt[10]{n^2+2}-\sqrt[10]{n^2+1}&\le n^{1/5}\left(\frac1{5n^2}-\frac1{10n^2+10}\right)\&=\frac1{10}n^{-9/5}\frac{n^2+2}{n^2+1}\end{align}$$ but also $$\begin{align}\sqrt[10]{n^2+2}-\sqrt[10]{n^2+1}&\ge n^{1/5}\left(\frac1{5n^2+10}-\frac1{10n^2}\right)\&=\frac1{10}n^{-9/5}\frac{n^2-2}{n^2+2}\end{align}$$ – robjohn Mar 27 '22 at 19:14
  • Wait a minute. Bernoullis inequality is this $(1+x)^n\geq (1+nx)$. But you did it the other way around and said $(1+\frac{1}{n^2})^{\frac{1}{10}}$$\leq 1+\frac{1}{10n^2}$... – Analysis_Mark Mar 27 '22 at 20:38
  • There are two cases for Bernoulli's Inequality. $1+ax\le(1+x)^a$ when $a\not\in(0,1)$ and $1+ax\ge(1+x)^a$ when $a\in[0,1]$. The difference boils down to the convexity or concavity of $x^a$. – robjohn Mar 27 '22 at 21:54
  • I give proofs at the ends of this answer (integer case) and this answer (rational case). The "standard" case is $1+nx\le(1+x)^n$ for $n\ge1$. To get the case for $a\in(0,1)$, just let $n=1/a$ and use the "standard" case. – robjohn Mar 27 '22 at 22:10
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If $n\to\infty$, that means $n$ is really, really big. $1$ or $2$ is basically negligible compared to $n^2$. So we can say $n^2+2\approx n$ and $n^2+1\approx n$, in which case the problem reduces to $\lim_{n\to\infty}n^{1/5}-n^{1/5}=0$

Raad Shaikh
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  • When $n$ is big, $n$ is negligible compared to $n^2$. What does that say about $\lim\limits_{n\to\infty}\left(\sqrt{n^2+n}-n\right)$? – robjohn Mar 27 '22 at 23:21
  • I would have naively thought this also becomes $0$, but I feel like you're trying to say that that's wrong and that a different approach is needed. Indeed, $\sqrt{n^2+n}-n=n(1+\frac{1}{n})^{1/2}-n\approx n(1+\frac{1}{2n}-1)=1/2$. (...unless this is also wrong?) Thank you for your example, I'll try to be more careful in the future! – Raad Shaikh Mar 28 '22 at 07:24
  • Indeed, the limit is $\frac12$. In fact $\frac12-\frac1{2n+2}\le\sqrt{n^2+n}-n\le\frac12$. – robjohn Mar 28 '22 at 08:37
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You can probably use the formula that generalises the conjugate : $a^m-b^m=(a-b)\sum_{k=0}^{m-1}a^k b^{m-1-k}$, with $m=10$, $a=\sqrt[10]{n^2+2}$ and $b=\sqrt[10]{n^2+1}$.

SacAndSac
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  • As it’s currently written, your answer is unclear. Please [edit] to add additional details that will help others understand how this addresses the question asked. You can find more information on how to write good answers in the help center. – Community Mar 26 '22 at 21:04
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Set $y:=n^2+1$, and consider $\lim y \rightarrow \infty$.

Then

$F(y):= (y+1)^{(1/10) }-y^{(1/10)}$

$=y^{(1/10) } \cdot$

$\left [\dfrac{(1+1/y)^{(1/10) }-1}{1/y}\right ] (1/y);$

$f:(x) =x^{(1/10) }$ and

$f'(x=1)=(1/10)x^{(-9/10)} _{x=1}$

$=1/10;$

$\lim_{y\rightarrow \infty}F(y)= $

$(\lim_{y \rightarrow \infty} y^{(-9/10)}) \cdot$

$(\lim_{y \rightarrow \infty} \dfrac{(1+1/y)^{(1/10)} -1}{1/y})=$

$0\cdot (1/10)=0$.

Peter Szilas
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