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Here, I got these limits without any problems. However, I am not confident to explain the differences. How can one explain the difference between two limits?

$$\lim \limits_{n \to \infty} x^{1/n}=1, x>0$$ and $$\lim \limits_{n \to \infty} {\frac {1}{2^n}}^{1/n}=\frac 1 2$$ Is it because $\frac 1 {2^n}\to0$ and thus contradicts $x>0$ condition above?

Bek Abdik
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    What $n$ goes to infinity in the first equation? – gebruiker May 27 '14 at 11:43
  • If everything is as written, then the first limit is $x^{1/2}$ if $x\geq 0$, and doesn't exist if $x < 0$ because $x^{1/2}$ isn't defined. The second is $0$. But maybe you left some things out? – GPerez May 27 '14 at 12:06
  • I am sorry for the confusion. Now, it's fixed – Bek Abdik May 27 '14 at 12:33
  • Well that's the entire point of the limit. You have an indeterminate expression and the tendency towards the limiting value pivotally depends on how the values converge to the limit. In this case you have one expression where the base of the power simply waits for the limiting process to happend and the other expression where the base is running away, resulting in effects of the base and exponent balance out at some value. That's an intuitive explanation. – orion May 27 '14 at 12:37
  • In both examples you are looking at the $n$th root, $n$ variable, of something. In the first example this something is constant, whereas in the second example it also depends on $n$. – Christian Blatter May 27 '14 at 12:40
  • "Is it because $\frac 1 {2^n}\to0$ and thus contradicts $x>0$ condition above?" Simple answer: yes – Ephraim May 27 '14 at 13:01

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In the first limit, you stated $x>0$. However in the second, $\lim_{n \to \infty}\frac{1}{2^n}=0$ so the difference between the two can bee seen as $$\lim \limits_{n \to \infty} x^{1/n}=1, x>0 \to x^0 = 1$$ vs $$\lim_{n \to \infty} \left({\frac {1}{2^n}}\right)^{1/n}\to 0^{^0}$$ Which is undefined. So we have to instead distribute the $1/n$ $$\lim_{n \to \infty} \left({\frac {1}{2^n}}\right)^{1/n} = \lim_{n \to \infty} \left({\frac {1^{1/n}}{2}}\right) \to \frac{1^0}{2} = \frac12$$ Simply put, $0^0 \neq 1$ which is why $\lim_{n \to \infty} \left({\frac {1}{2^n}}\right)^{1/n} \neq 1$

Ephraim
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