$\lim_{x\to\infty }\frac{q^{x}}{x!}$ I need to find a limit of this sequence and prove it using the definition.  I know that we can have several variants of the limit of $q^{x}$: if $1&gt;q&gt;-1$ the limit is $0$, for $q=1$ the limit is $1$, for $q&gt;1$ the limit is $\infty $,  and for $q\leq -1$ it doesn't exist.  The limit of factorial $x!$ should be $\infty $, but I'm not sure and I don't know how to prove it.  Can you help me please with the whole fraction.  I don't know how to find a limit if we have something like $\frac{0}{\infty }$ or $\frac{1}{\infty }$ or $\frac{\infty}{\infty }$.
First I tried to use integration: $$y=\lim_{n\to\infty}\frac{a^n}{n!}=\lim_{n\to\infty}\frac{a}{1}\cdot\frac{a}{2}\cdot\frac{a}{3}\cdots\frac{a}{n}$$ $$\log y=\lim_{n\to\infty}\sum_{r=1}^n\log\frac{a}{r}$$ But I could not express it as a riemann integral. Now I am thinking about sandwich theorem.  $$\frac{a}{n!}=\frac{a}{1}\cdot\frac{a}{2}\cdot\frac{a}{3}\cdots\frac{a}{t} \cdot\frac{a}{t+1}\cdot\frac{a}{t+2}\cdots\frac{a}{n}=\frac{a}{t!}\cdot\frac{a}{t+1}\cdot\frac{a}{t+2}\cdots\frac{a}{n}$$ Since $\frac{a}{t+1}&gt;\frac{a}{t+2}&gt;\frac{a}{t+1}&gt;\cdots&gt;\frac{a}{n}$ $$\frac{a^n}{n!}&lt;\frac{a^t}{t!}\cdot\big(\frac{a}{t+1}\big)^{n-t}$$ since $\frac{a}{t+1}&lt;1$, $$\lim_{n\to\infty}\big(\frac{a}{t+1}\big)^{n-t}=0$$ Hence, $$\lim_{n\to\infty}\frac{a^t}{t!}\big(\frac{a}{t+1}\big)^{n-t}=0$$ And by using sandwich theorem, $y=0$. Is this correct?
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