Does $\lim_{x\to\infty}\left( (t+x)^{\frac{1}{n}}+(t-x)^{\frac{1}{n}}\right)$ exist?

Question

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 $L_{1,3}= {(-1)}^{\frac{1}{6}}.\infty$

Answer 1

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.

Answer 2

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 $2t.$