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Prove $$\lim_{n\to\infty} x_n^{1/k} = \left(\lim_{n\to\infty} x_n\right)^{1/k}.$$

$\lim x_n = x$, and $x_n \ge 0$, $\varepsilon >0$

  1. when $x=0$, $|x_n-x| \lt \varepsilon^k$ $\rightarrow$ $|x_n|< \varepsilon^k$ $\rightarrow$ $x_n^{\frac 1k} \lt \varepsilon$.

  2. when $x \gt 0$, $|x_n^{\frac 1k}-x^{\frac 1k}|= \dfrac {|x_n-x|}{x_n^\frac{k-1}{k}+x^\frac{k-1}{k}}\le \dfrac {|x_n-x|}{x^\frac{k-1}{k}} $. Suppose $|x_n-x| \lt \varepsilon\cdot x^\frac{k-1}{k}$. Then, $|x_n^{\frac 1k}-x^{\frac 1k}|<\varepsilon$.

Could you tell me whether the proof is valid??

Thank you in advance.

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    $\begingroup$ Sorry but the identity $$|a^{1/k}-b^{1/k}|=\frac{|a-b|}{a^{1-1/k}+b^{1-1/k}}$$ which you take for granted, is pure fantasy when $k\ne2$. $\endgroup$
    – Did
    Mar 10, 2018 at 9:13
  • $\begingroup$ The inequality $(x+y)^{1/k} \le x^{1/k} + y^{1/k}$ for $k>1$, $x,y>0$ might be helpful. $\endgroup$
    – Hanul Jeon
    Mar 11, 2018 at 7:47
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    $\begingroup$ Don't use \frac in exponents or limits of integrals. I have changed the formatting of the title so as to make it take up less vertical space -- this is a policy to ensure that the scarce space on the main page is distributed evenly over the questions. See here for more information. Please take this into consideration for future questions. Thanks in advance. $\endgroup$
    – GNUSupporter 8964民主女神 地下教會
    Mar 12, 2018 at 17:12
  • $\begingroup$ Are you sure you mean $\lim_{x \to \infty}$, or do you really mean $\lim_{n \to \infty}$? And have you learned the general fact that if $f$ is a continuous function, then $\lim_{n \to \infty} f(x_n) = f(\lim_{n \to \infty} x_n)$? It may be algebraically neater to prove this general fact and then apply it to the case $f(x) = x^{1/k}$. $\endgroup$
    – Connor Harris
    Mar 12, 2018 at 17:24

1 Answer 1

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Your $x=0$ is fine as long as you make some mention of "there exists some $N$ so that for all $n\geq N$, $|x_n-x|<\epsilon^k$.

To deal with $x > 0$ case, let $y_n = x_n^{1/k}$, $x=y^{1/k}$, then

$$y_n^k - y^k = (y_n-y)(y_n^{k-1}+y_n^{k-2}y + \dotsb + y^{k-1})$$ Rearranging and rewriting in terms of $x_n$, and noting that all terms are non-negative

$$|x_n^{1/k}-x^{1/k}| = \frac{|x_n-x |}{x_n^{1-1/k} + x_n^{1-2/k}x^{1/k} + \dotsb + x_n^{1/k}x^{1-2/k} + x^{1-1/k}}$$ Now, fix any $m$ that satisfies $0<m<x$. Given that $x_n\to x$, it follows that there must exist some $N_1$ so that $m < x_n$ for all $n\geq N_1$ (as $x_n$ eventually gets arbitrarily close to $x$).

Pick $\epsilon > 0$, as $x_n \to x$ there exists some $N_2$ so that for $n\geq N_2$, $$|x_n-x| \leq \epsilon m^{1-1/k} k$$ hence if $n \geq \max\{N_1,N_2\}$ then

$$|x_n^{1/k}-x^{1/k}| < \frac{|x_n-x|}{m^{1-1/k}+m^{1-2/k}m^{1/k}+\dotsb + m^{1-1/k}} = \frac{|x_n-x|}{km^{1-1/k}} < \epsilon $$

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  • $\begingroup$ Could you tell me how to do this "$y_n^k - y^k = (y_n-y)(y_n^{k-1} +y_n^{k-2}y+...+y^{k-1})$"?? $\endgroup$
    – shk910
    Mar 12, 2018 at 22:57
  • $\begingroup$ In summation notation, changing to $a$ and $b$, expand and then remove the last and first term from the sums respectively, i.e., \begin{align} (a-b)\sum_{i=0}^{k-1} a^i b^{k-1-i} &= \sum_{i=0}^{k-1} a^{(i)+1} b^{k-1-i} - \sum_{i=0}^{k-1} a^i b^{(k-1-i)+1}\\ &= \sum_{i=0}^{k-2} a^{i+1}b^{k-1-i} + a^{(k-1)+1} - \sum_{i=1}^{k-1} a^ib^{k-i} - b^{k-(0)}\\ &= a^k - b^k + \underbrace{\sum_{i'=1}^{k-1} a^{i'}b^{k-i'} - \sum_{i=1}^{k-1} a^ib^{k-i}}_{=\;0}\\ &= a^k-b^k \end{align} where we substituted $i'=i+1$ in the first sum. $$ $\endgroup$
    – adfriedman
    Mar 12, 2018 at 23:29
  • $\begingroup$ Alternatively, use polynomial division to compute $$\frac{a^k-b^k}{a-b}$$ directly. $\endgroup$
    – adfriedman
    Mar 12, 2018 at 23:32
  • $\begingroup$ I got it. Thanks for detailed explanation !! $\endgroup$
    – shk910
    Mar 12, 2018 at 23:49
  • $\begingroup$ I am sorry. Do you mind if I ask you to show how to rearrange in terms of $x_n$ ? For example, just one $y_n^{k-2}y$ to $\frac 1{x_n^{1- \frac 1k}} x^{\frac 1k}$ $\endgroup$
    – shk910
    Mar 13, 2018 at 0:32

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