Function $v_n$ takes the values of the binomial coefficients as inputs: $v_1=m,\; v_2=\frac{m(m-1)}{1\cdot2},\; ..., v_n=\frac{m(m-1)(m-2)...(m-n+1)}{1\cdot2\cdot3\cdot...\cdot n},..., \;$ where $m$ is a positive integer.
Find$$\lim_{n\to\infty} v_n$$
source: Berman. G.N. A Problem Book in Mathematical Analysis, Moscow, 1967.
You simply have to notice that the terms of the sequence become $0$ from some point on. When $n>m$, you have $$v_n=\frac{m(m-1)(m-2)\cdots(m-m)\cdots(m-n+1)}{1\cdot2\cdot3\cdots n}=0$$ Using binomial coefficients, this is saying that ${m\choose n}=0$ for $n>m$. Since the sequence consists of just zeros eventually, you have $$\lim_{n\to\infty}v_n=0 $$