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Let $x_n \geq -1$ for all n $\in \mathbb{N}$ and $ \lim_{n \to \infty}(x_n)=0 $. Show that If $p \in \mathbb{N}$, then $\lim_{n \to \infty}\sqrt[p]{1+x_n}=1 $

Suggestion of how to do it, please.

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  • $\begingroup$ you mention $p \in \mathbb N$ once but then do nothing with it. $\endgroup$
    – Will Jagy
    Sep 22, 2018 at 1:03
  • $\begingroup$ I'm sorry, I already correct it $\endgroup$
    – VERA
    Sep 22, 2018 at 1:05
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    $\begingroup$ One approach is to simply note the continuity of the map $x \mapsto \sqrt[p]{x}$, which trivializes this proof. Could you give us a bit more context? What class is this for? What tools do you have at your disposal? What kind of proof are you expecting here? $\endgroup$ Sep 22, 2018 at 1:11
  • $\begingroup$ this is analysis 1 and I'm starting the whole theme of the sequences and we have the sequence definition and all the basic elements of the theme, but I do not know how to approach this exercise. $\endgroup$
    – VERA
    Sep 22, 2018 at 1:16
  • $\begingroup$ sequence convergent $\endgroup$
    – VERA
    Sep 22, 2018 at 1:21

3 Answers 3

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Here's another approach. Since $x_n>-1$ and $0<1/p\leq1$, you can use Bernoulli's inequality: $$(1+x_n)^{1/p}\leq1+\frac{x_n}{p}$$ If $x_n\geq0$, one obtains $$1\leq(1+x_n)^{1/p}\leq 1+x_n/p$$ On the other hand, if $-1<x_n<0$, one has $$1-|x_n|<(1-|x_n|)^{1/p}\leq1-|x_n|/p$$ Consequently, one takes the limit $n\rightarrow\infty$ and obtains 1.

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Your approach should focus on finding how large $n$ needs to be (in terms of $\epsilon$) to guarantee $|\sqrt[p]{1 + x_n} - 1| < \epsilon$. In this case, you'll need to use the fact that you can always find $n$ large enough to guarantee $|x_n| < \epsilon$. You should be able to use the fact that $\sqrt[p]{x} - \sqrt[p]{y} \leq \sqrt[p]{x - y}$ when $x > y$ to get $|\sqrt[p]{1 + x_n} - 1| < \sqrt[p]{|x_n|}$. See real analysis - Difference of nth roots vs nth root of difference ... for the general statement and real analysis - Prove the difference of roots is less than or ... for the statement specifically with square roots (and absolute values!) and see if that works for you.

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  • $\begingroup$ But my question is, how can I manipulate the radiation to get there?. I have not idea $\endgroup$
    – VERA
    Sep 22, 2018 at 2:15
  • $\begingroup$ I will add to my answer $\endgroup$ Sep 22, 2018 at 2:16
  • $\begingroup$ and that's because $\endgroup$
    – VERA
    Sep 22, 2018 at 2:41
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If you can show that $f(x)=x^{1/p}$ is continuous then my proof will work:

Remember that if a function $f$ is continuous then $\lim_{x\to x_0} f(x)=f(x_0)=f(\lim_{x\to x_0} x)$ or written in an equivalent way this is just $\lim_{n\to \infty} f(x_n) = f(x) = f(\lim_{n\to \infty} x_n)$. But because $f(x)=x^{1/p}$ is continuous it follows from these facts that $\lim_{n\to \infty} (1+x_n)^{1/p}= (1+\lim_{n\to \infty} x_n)^{1/p} = (1+0)^{1/p}=1$

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