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$\lim_\limits{n\to \infty} \frac{1}{n!}$

I really don't know how to solve this. I usually use this theorem to solve limits of oscillating functions such as sine and cosine. But for this function I only start from the premise that it will always be greater than zero.

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  • $\begingroup$ I made some improvements to the MathJax in your question (which you can view by clicking on "edited x ago"). The main one is that you should use \infty to get the infinity symbol. $\endgroup$ – Rhys Steele May 7 '18 at 14:17
  • $\begingroup$ thank you. I am very bad at writing here. $\endgroup$ – Virginia Martín Herrera May 7 '18 at 14:18
  • $\begingroup$ Do you HAVE to use the sandwich lemma? Because you can say straightaway that $\frac{1}{n!}\to 0$ as $n\to\infty$ $\endgroup$ – Michal Dvořák May 7 '18 at 14:22
  • $\begingroup$ Yes I have to. I know that the limit of the function when n tends to infinity is 0 but I have to explain it using the theorem. That's why I asked for help. $\endgroup$ – Virginia Martín Herrera May 7 '18 at 14:27
  • $\begingroup$ @MichalDvořák There are lots of limits that you can say straightaway are $0$. However when you're at the level of being asked this exercise you're probably expected to give justifications from basic results you know (e.g. showing $\frac1n \to 0$ is an application of the Archimedean property of $\mathbb{R}$). Admittedly it's been a long time since I did any of this stuff but it seems like a justification without squeeze theorem would be longer and would probably essentially do the same kind of thing anyway? $\endgroup$ – Rhys Steele May 7 '18 at 14:31
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As you guessed, $\forall n\in \mathbb{N}$:

$$ 0< \frac{1}{n!} $$

But you could also use:

$$ \frac{1}{n!}\leq\frac{1}{n} $$

Since $\frac{1}{(n-1)!}\leq 1$ for $n\in \mathbb{N}^*$.

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$$\frac{1}{n}\geq\frac{1}{n!}>0, \forall n \in \mathbb R^+$$ and $$\lim_{x\rightarrow\infty}\frac{1}{n}=\lim_{x\rightarrow\infty}0=0$$ so $$\lim_{x\rightarrow\infty} \frac{1}{n!}=0$$ by squeeze theorem.

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