Is $\{f(x_n^2)\}$ a Cauchy sequence if $|x_n-x_{n+1}|<\frac{1}{10^n}$ and $f$ is continuous?

Question

Let $\{x_n\}$ be a sequence in $\Bbb R$ that satisfies $$|x_n-x_{n+1}|<\frac{1}{10^n}\tag{$*$}$$ for each $n$, and let $f:\Bbb R\to\Bbb R$ be continuous. I'd like to determine whether $\{f(x_n^2)\}$ is a Cauchy sequence.

It occurred to me that continuity preserves sequential limits, so I decide to consider the continuous function $f\circ g$ with $g:\Bbb R\to\Bbb R$ defined by $g(x)=x^2$. Now $\{x_n\}$ is a Cauchy sequence in $\Bbb R$, which can be easily verified by an application of ($*$). Therefore, $\{x_n\}$ is a convergent sequence, and so is $\{(f\circ g)(x_n)\}$, which in turn tells us that $\{f(x_n^2)\}$ is a Cauchy sequence. Is my approach correct? Thank you.

Answer 1

In general, the image of a Cauchy sequence under a continuous function is not a Cauchy sequence. But here you can use that continuous functions preserve sequential limits and that the domain of the function is complete: The Cauchy sequence converges to some point $x_*$ in the domain of the function; now you can use that the function is continuous at $x_*$, and therefore the image of the sequence converges to the image of $x_*$ and is therefore a Cauchy sequence.

Answer 2

$f:\mathbb{R}\to \mathbb{R} $ continuous.

$(x_n) $ cauchy implies $(x_n) $ convergent due to completeness of $\mathbb{R}$

$(x_n) $ convergent implies $(x_n) ^2 $ convergent.

A continuous function maps convergent sequence to convergent sequence.

$f((x_n) ^2) $ is also convergent.

And every convergent sequence is cauchy.

Implies, $f((x_n) ^2) $ is cauchy.