# A limit of a sequence problem

• Mar 10th 2010, 12:50 PM
Runty
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?
• Mar 10th 2010, 12:58 PM
Plato
If you assume that $L>M$ then $\epsilon=\frac{L-M}{2}>0$ makes sense.
• Mar 10th 2010, 01:14 PM
Runty
Quote:

Originally Posted by Plato
If you assume that $L>M$ then $\epsilon=\frac{L-M}{2}>0$ makes sense.

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?
• Mar 10th 2010, 01:43 PM
Plato
Quote:

Originally Posted by Runty
I'm afraid I don't fully understand how these things can be applied. How exactly would one show this?

$\left| {a_n - L} \right| < \frac{{L - M}}
{2} \Rightarrow M < \frac{{L + M}}
{2} < a_n$