<s>[INST] How to calculate limit of a function having factorial in denominator For $n$ tending to infinity find the following limit
$$\frac{2^n}{n!}.$$
I have a feeling that it is multiplication of many numbers with the last one turning to $0$ but the first one is finite so limit should be $0$. But I am not sure and neither am I able to put it in mathematical form.
Thank you for your help.
 [/INST] $$0<\frac{2^n}{n!}=\frac{2}{1} \cdot \frac{2}{2} \cdot \frac{2}{3} \cdot \frac{2}{4} \cdot \dots \cdot \frac{2}{ n} \leq \frac{2}{1} \cdot \frac{2}{2} \cdot \frac{2}{3} \cdot \frac{2}{3} \cdot \dots \cdot \frac{2}{3}=\frac{2}{1} \cdot \frac{2}{2} \cdot \left (\frac{2}{3} \right )^{n-2}=2 \left ( \frac{2}{3} \right )^{n-2}$$
As $n \rightarrow \infty$, $\left ( \frac{2}{3} \right )^{n-2} \rightarrow 0$
Therefore, from the Squeeze Theorem $$\lim_{n \rightarrow \infty} \frac{2^n}{n!}=0$$
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