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 $\frac{a}{t+1}>\frac{a}{t+2}>\frac{a}{t+1}>\cdots>\frac{a}{n}$ $$\frac{a^n}{n!}<\frac{a^t}{t!}\cdot\big(\frac{a}{t+1}\big)^{n-t}$$ since $\frac{a}{t+1}<1$, $$\lim_{n\to\infty}\big(\frac{a}{t+1}\big)^{n-t}=0$$ 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?