# Thread: A limit of a sequence problem

1. ## A limit of a sequence problem

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

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 $\displaystyle L>M$ then $\displaystyle \epsilon=\frac{L-M}{2}>0$ makes sense.
If you assume that $\displaystyle L>M$ then $\displaystyle \epsilon=\frac{L-M}{2}>0$ makes sense.
$\displaystyle \left| {a_n - L} \right| < \frac{{L - M}} {2} \Rightarrow M < \frac{{L + M}} {2} < a_n$ that is a contradiction.