# Math Help - A limit of a sequence problem

1. ## A limit of a sequence problem

Suppose that $a_n\rightarrow L$. Show that if $a_n\leq M$ for all $n$, then $L\leq M$.
Prove by contradiction and by choosing $\epsilon =\frac{L-M}{2}$.

I have a standard proof for how to solve this, but not a contradictory proof (which is what I need). Can someone help with that?

2. If you assume that $L>M$ then $\epsilon=\frac{L-M}{2}>0$ makes sense.
That is by contradiction.

3. Originally Posted by Plato
If you assume that $L>M$ then $\epsilon=\frac{L-M}{2}>0$ makes sense.
That is by contradiction.
I'm afraid I don't fully understand how these things can be applied. The answer sounds right, but I can't slot it into my proof correctly. How exactly would one show this?

4. Originally Posted by Runty
I'm afraid I don't fully understand how these things can be applied. How exactly would one show this?
$\left| {a_n - L} \right| < \frac{{L - M}}
{2} \Rightarrow M < \frac{{L + M}}
{2} < a_n$
that is a contradiction.