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Suppose that $a \in (l, r)$ and $f : (l, r) \smallsetminus \{a\} \to \mathbb{R}$ satisfies $f(x) \geq 0$ for all $x \in (l, r) \smallsetminus \{a\}$. Prove that if $\lim_{x\to a} f(x)$ exists then the limit is non-negative.

Solution

Because $f(x)$ is always non negative, the limit can never be negative, therefore it must be positive. The problem is that this is not really a satisfactory proof, and I need to use the epsilon delta limit definition, can anyone help here? Thank you!

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  • $\begingroup$ Take an $l < 0$. Can you find an $\epsilon > 0$ that shows that $l \neq \lim_{x\to a} f(x)$? $\endgroup$ – Daniel Fischer Oct 16 '14 at 18:31
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The usual route is by contradiction.

Suppose to the contrary that $\lim_{x \to a} f(x) = L < 0$. Let $\epsilon = -L/2 > 0$. There exists $\delta > 0$ with the property that $x \in (l,r) \setminus \{a\}$ and $|x-a| < \delta$ implies $|f(x) - L| < -L/2$. This implies in turn that if $x \in (l,r) \setminus \{a\}$ and $|x-a| < \delta$ then $f(x) < L/2$. But $L/2 < 0$, in contradiction with your assumption that $f(x) \ge 0$.

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