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I could use some help on a proof. Suppose $f(x)≥0$ for all $x$ (except perhaps at $x=a$). Show that if $\lim\limits_{x→a}f(x)=L$, then $L≥0$. Any guidance would be helpful.

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Do you know how to argue by contradiction? – Pedro Tamaroff Sep 27 '12 at 19:28
Sometimes. I had a feeling that it was going to be necessary. – user42864 Sep 27 '12 at 19:30
up vote 1 down vote accepted

Assume that $L < 0$. Then there is an $\delta > 0$ such that if $\lvert f(x) - L\lvert < \frac{-L}{2}$ whenever $\lvert x - a\lvert < \delta$. Now consider: $$ \begin{align} \lvert f(x) - L\lvert &< \frac{-L}{2} & \Rightarrow \\ \frac{L}{2} < f(x) - L &< \frac{-L}{2} & \Rightarrow \\ \frac{3L}{2} < f(x) &< \frac{L}{2}. \end{align} $$ So for all $x$ for which $\lvert x - a\lvert < \delta$ you have $$ \frac{3L}{2} < f(x) < \frac{L}{2}. $$ Can you finish the argument from here?

Edit: To add a bit more explanation here. What we am doing above is trying to prove the statement by assuming the the negation of the statement. I.e. we assume that what we want to prove is false. Then we want to prove that the assumption is false. That is why we start out by assuming that $L<0$. What we want to prove is that there is some $x$ such that $f(x) < 0$. Above then we get to the equation that for $x$ in the interval $(x-a, x+a)\setminus \{a\}$ we have $$ \frac{3L}{2} < f(x) < \frac{L}{2} <0. $$ But that tells us exactly that for any $x \in (x-a, x+a)\setminus \{a\}$, $f(x) < 0$.

We say that we have proved the statement by contradiction.

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Got it! Thanks! – user42864 Sep 27 '12 at 19:37
@user42864: No problem. Glad to help. (Remember to accept one of the answers. I noticed that you have also asked another question without accepting the answer given. In general you will be more successful in getting answers if you accept one of the answers given.) – Thomas Sep 27 '12 at 19:39
I'm a newbie, as of today. Thank you for letting me know. – user42864 Sep 27 '12 at 19:54
For what it its worth, I didn't like the suggested proofs. I am still pushing through the proof. – user42864 Sep 27 '12 at 19:55
@user42864: I will edit and update my answer with the last bit. – Thomas Sep 27 '12 at 20:38

Suppose not. Then we can find $\delta $ such that $|f(x) - L| = f(x) - L < - L$ for $0 <|x-a | <\delta $ - that is, $f(x) <0,$ but this is absurd.

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The hint in the problem states to let ϵ = L/2. I have no idea what to do with that – user42864 Sep 27 '12 at 19:34

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