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Let $(x_k)_k\subseteq\mathbb{R}$ be a Cauchy sequence. Is $(e^{1/x_k})_k$ a Cauchy sequence as well? I am not sure how to approach this question.

Tommy1234
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tnura23
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1 Answers1

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Since you are in $\mathbb{R}$, a sequence converges iff the sequence is Cauchy. So since $(x_k)_k$ is Cauchy, $(x_k)_k$ converges to some $x\in\mathbb{R}$.

  • If $x\neq 0$, by continuity, $(e^{1/x_k})_k$ converges to $e^{1/x}$, so that $(e^{1/x_k})_k$ is Cauchy.
  • If $x=0$, $1/x_k\,\rightarrow_k\infty$, so there exists $N\in \mathbb{N}$ and $\varepsilon >0$ such that for all $k\geq N$, $$|x_{k+1}-x_k|\geq \varepsilon,$$ so that $(1/x_k)_k$ is not Cauchy and therefore neither $(e^{1/x_k})_k$.
SacAndSac
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    Unless $x_k\to 0$, for example for $x_k=\frac{1}{k}$ we get the sequence ${e^k}$ which is clearly not Cauchy. – Tommy1234 Apr 05 '22 at 17:38
  • But what if lets say x_k=1/n for every n in natural numbers which is convergent so it is Cauchy. Then wouldn't that make the sequence of e^1/x_k to e^n which in turn wouldn't be convergent hence, not Cauchy? – tnura23 Apr 05 '22 at 17:40
  • @Tommy1234 yes indeed – SacAndSac Apr 05 '22 at 18:00