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I and my friend were sitting in the maths lesson and we suddenly came up with what infinite root out of x is equal to. And we came up with this this equation:

$$ \sqrt[\infty]{x} = 1+ (x-1)×10^{-\infty} $$

I wonder if this makes any sense. (Im 10th former, and sorry if this is an out of topic question)

Antonio Vargas
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yamahamm
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2 Answers2

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Let $x>0$. Write it as

$$ \displaystyle\lim_{n\to \infty}\sqrt[n]{x}=\lim_{n\to \infty} x^{1/n} $$ You see that the power $1/n$ goes to $0$ when $n$ increases so $x^{1/n}$ goes to $x^0$ which is $1$.

draks ...
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$\sqrt[\infty]{x}$ can be interpreted as

$\lim\limits_{a\rightarrow\infty}\sqrt[a]{x}$

and $1+(x-1)\cdot 10^{-\infty}$ can be interpreted as

$\lim\limits_{a\rightarrow\infty}(1+\large\frac{x-1}{10^a})$

In this sense the given equation is true since

$\lim\limits_{a\rightarrow\infty}\sqrt[a]{x}= \lim\limits_{a\rightarrow\infty}(1+\large\frac{x-1}{10^a})=1$

Danny
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  • But does this work with numbers bigger than one? – yamahamm Apr 02 '14 at 20:31
  • What exactly do you mean by numbers bigger than one? The "infinite root" of ANY (positive) number is 1. – Eff Apr 02 '14 at 20:33
  • This equation holds for all $x$ bigger than zero.. – Danny Apr 02 '14 at 20:33
  • @Danny so you mean that with any number bigger than 0 equation is equal to 1, or this formula has some sense? – yamahamm Apr 02 '14 at 20:39
  • Yes. Both sides of your equation are equal to 1 for $x>0$. The sense of your equation is another question.. it is not difficult to construct equations like yours, but maybe there is an interesting interpretation.. :) – Danny Apr 02 '14 at 20:42
  • Ok, thats all i wanted to know, thanks :) – yamahamm Apr 02 '14 at 20:44