1
$\begingroup$

Given that $t\in\mathbb R$ and $n$ is a positive odd integer , does the following limit exist ?

$$L_{t,n}=\lim_{x\to\infty}\left( (t+x)^{\frac{1}{n}}+(t-x)^{\frac{1}{n}}\right)$$

For $t=0$ ,

$$\lim_{x\to\infty}\left( (x)^{\frac{1}{n}}+(-x)^{\frac{1}{n}}\right)= \lim_{x\to\infty}\left( (x)^{\frac{1}{n}}-(x)^{\frac{1}{n}}\right)=0$$

(Because $n$ is a positive odd integer , we can take the negative sign out)

Now , my speculation is that $t$ , whatever it's value be, being just a finite real number , it's effect should be decreasing as $x$ increases. And as $x\to\infty$ , the effect of $t$ will be negligible. Hence , the limit should be equal to zero $\forall t\in\mathbb R$

However , online calculators are saying that limit does not exist. And Wolfram alpha is giving weird results like $L_{1,3}= {(-1)}^{\frac{1}{6}}.\infty$

$\endgroup$
4
  • $\begingroup$ I think the $t=0$ exact case was more of a lucky cancellation that collapses the entire problem for all values of $x$, rather than some "leading order" dominant behaviour. Have you tried a series expansion to test your intuition? $\endgroup$ Commented Sep 6, 2023 at 19:02
  • $\begingroup$ Also I have no access to pen and paper at the moment but it very much looks like Bernoulli's inequality may be of use, if you factor out $t$ $\endgroup$ Commented Sep 6, 2023 at 19:04
  • $\begingroup$ @TheoDiamantakis Yes, I was exactly thinking about that "leading term" dominance over finite $t$. Unfortunately, I haven't learn those types of series yet as I only know basic calculus. Thanks ! $\endgroup$ Commented Sep 6, 2023 at 19:04
  • $\begingroup$ Just because $t$ is small and finite as you limit $x$ doesn't mean it can't necessarily wreak havoc on this type of limit. I'm sure you'll agree that $\lim_{x\rightarrow \infty} (x+t)^n - x^n = 0 $ if $ t=0$, but never any other time! $\endgroup$ Commented Sep 6, 2023 at 19:10

2 Answers 2

3
$\begingroup$

Since n is an odd natural number, we have: $$\sqrt[n]{t+x} + \sqrt[n]{t-x}= \sqrt[n]{x+t}- \sqrt[n]{x-t}$$ by the Identity $$a^n - b^n = (a-b)(a^{n-1} + a^{n-2}*b +... + b^{n-1})$$ put $a= \sqrt[n]{x+t}$, and $b= \sqrt[n]{x-t}$.
Now, you want to multiply and divide the expression by $(a^{n-1} + a^{n-2}*b +... + b^{n-1})$. Then, you get $$\sqrt[n]{x+t}-\sqrt[n]{x-t} = \frac{2t}{(a^{n-1} + a^{n-2}*b +... + b^{n-1})}$$ It is now very easy to see why your claim is true.

$\endgroup$
7
  • $\begingroup$ I think if you wrote this out a bit, using double dollar signs etc, it would improve the quality of this post. $\endgroup$ Commented Sep 6, 2023 at 19:38
  • $\begingroup$ Thank you for the tip! $\endgroup$
    – Arfin
    Commented Sep 6, 2023 at 19:41
  • $\begingroup$ Then why is Wolfram alpha giving weird results and other calculators saying that limit does not exist ? Thanks for the answer. $\endgroup$ Commented Sep 7, 2023 at 7:14
  • $\begingroup$ Maybe, because you have misinput something? Try special cases by assigning values to $n$ and $t$. Maybe, you forgot to state that $n$ is odd in your program? It would help if you provided your code after all. @An_Elephant $\endgroup$
    – Arfin
    Commented Sep 7, 2023 at 7:29
  • $\begingroup$ Please see question. As said, it gives wrong result for $L_{1,3}$ . Here is link : wolframalpha.com/… $\endgroup$ Commented Sep 7, 2023 at 7:51
2
$\begingroup$

This explanation is more costly than the answer by @Aria as the derivative, namely MVT, is applied. For $x>t$ we get $$0\le (t+x)^{1/n}+(t-x)^{1/n}\\ =(x+t)^{1/n}-(x-t)^{1/n}\\ ={2t\over n}u^{1/n-1}\le {2t\over n}(x-t)^{1/n-1}$$ for some $u,$ $x-t<u<x+t.$ The last expression tends to $0$ when $x\to \infty,$ because the exponent is negative for $n>1.$

For $n=1$ the expression is constantly equal $2t.$

$\endgroup$
1
  • $\begingroup$ Then why is Wolfram alpha giving weird results and other calculators saying that limit does not exist ? Thanks for the answer. $\endgroup$ Commented Sep 7, 2023 at 7:14

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .