Show that if $x_n ≥ -1$ for all n $\in \mathbb{N}$ and $ \lim_{n \to \infty}(x_n)=0 $

Question

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.

Answer 1

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.

Answer 2

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.

Answer 3

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 $\lim_{n\to \infty} (1+x_n)^{1/p}= (1+\lim_{n\to \infty} x_n)^{1/p} = (1+0)^{1/p}=1$