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

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 $\min\{a,b\}$.

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You can also take $c$ between $a$ and $b$ (if they are equal then the result is trivial) and then from the limits, you know after a given $n_0$, all the $a_n$ will be on one side of $c$ and all the $b_n$ will be on the other side. So after that $n_0$, all $a_n$ will be smaller than the $c$ that is smaller than all the $b_n$ (or the same thing where you swap $a_n$ and $b_n$) so your sequence of the minimums will be equal to $a_n$ after this given $n_0$ so its limit is the minimum of $a$ and $b$. – xavierm02 Dec 16 '12 at 19:09
^ The above should be its own answer and not just a comment. – Joe Z. Dec 19 '12 at 0:17

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