# Let $b, L \in \Bbb R$. If $b \ge L - \varepsilon$, then $b \ge L$

Let $b, L \in \Bbb R$. Prove that if $b \ge L - \varepsilon$ for all positive $\varepsilon$, then $b \ge L$

I started of by assuming the "if" part. But, are you supposed to use cases to prove this? Is there an epsilon you can pick? Not sure what the approach should be here.

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Suppose $b\geq L-\epsilon$ for all $\epsilon>0$.
If on the contrary $b<L$, we may set $\epsilon=\frac{L-b}{2}>0$. Then $L-\epsilon=L-\frac{L-b}{2}=\frac{L+b}{2}>b$, a contradiction.
Using the contradiction method, it's still a little unclear to me. So I substituted $\frac{L-b}{2}$ for $\epsilon$, which gives me $b \ge \frac{L-b}{2}$. But, I can't seem to wrap my around the reason this becomes a contradiction. –  icanc Dec 7 '12 at 1:26
@JasperLoy It makes sense now since we assumed that $b \ge L - \epsilon$ and $b \lt L$. Thank you! –  icanc Dec 7 '12 at 1:46