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I am trying to show that if $f(x) \geq 0$ for every $x \in (-\infty, a)$ and $ \lim_{x \rightarrow a^-} f(x)$ exists then $\lim_{x \rightarrow a^-} f(x) \geq 0$. Even though it is intuitively obvious, the proof I have come up with is so easy it concerns me that I'm missing something:

Suppose that $ \lim_{x \rightarrow a^-} f(x) = L < 0$. This means that for every $\epsilon > 0$ there exists a $\delta > 0$ such that $|x - a| < \delta \implies |f(x) - L| < \epsilon$ and $x < a$. Since $L < 0$ and $f(x) \geq 0$, $f(x) - L > 0$ so $0 < f(x) - L < \epsilon$. Now, choose $\epsilon = -L$. Then

$ 0 < f(x) -L < -L \implies 0 < f(x) < 0 $

which is a contradiction, hence, $L\geq 0$

Does this look right?

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    This looks correct. Of course for the left-sided limit |x-a|<\delta can be simplified by removing the modulus... You can clean it up to make it better, but this is fine the way it is.2011-06-28

1 Answers 1

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For the sake of having an answer:

  1. Your argument is fine except for the slight glitch pointed out by Dylan Moreland: At the very end you can only conclude $0 \leq f(x) \lt 0$, which is of course a contradiction, too. As for the "simplification" resulting from $x \lt a$ allowing you to write $0 \lt a - x \lt \delta$, this seems rather immaterial to me.
  2. The limit point $L$ must be in the closure of $[0,\infty)$, hence $L \geq 0$, so no proof by contradiction is needed here.