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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, we may set $\epsilon=\frac{L-b}{2}>0$. Then $L-\epsilon=L-\frac{L-b}{2}=\frac{L+b}{2}>b$, a contradiction.

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    I've run out of answers to upvote :(2012-12-30