Show $\lim_{n \to \infty} \min\{a_{n},b_{n}\} = \min\{a,b\}$

Question

If $\lim_{n \to \infty} a_{n} = a$ and $\lim_{n \to \infty} b_{n} = b$, how can we show that $\lim_{n \to \infty} \min\{a_{n},b_{n}\} = \min\{a,b\}$?

I say $\min\{a_{n},b_{n}\} $ has two cases: $a_{n}$ and $b_{n}$. So (1) $\lim_{n \to \infty} a_{n} = a$ and (2) $\lim_{n \to \infty} b_{n} = b$. Now I don't know how to imply the $\min\{a,b\}$.

Answer 1

Hint: $$\max\left\{a_n,b_n\right\}=\frac{1}{2}(a_n+b_n)+\frac{1}{2}\left|a_n-b_n\right|$$ and $$\min\left\{a_n,b_n\right\}=\frac{1}{2}(a_n+b_n)-\frac{1}{2}\left|a_n-b_n\right|$$

Answer 2

If $a_n\to a$ and $b_n\to b$, then $$\min (a_n,b_n)=\frac{1}{2}(a_n+b_n-|a_n-b_n|)\to \frac{1}{2}(a+b-|a-b|)=\min (a,b).$$