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Help me out here, people. I have a proof that I, for some reason, can not figure out.

Let $x_1$ and $x_2$ be any numbers that both satisfy $0<|x-a|<\delta$. Show that $|f(x_1) – f(x_2)|< \varepsilon$. I think that I have to use the triangle inequality, but I am not exactly sure.

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Are we allowed to choose the function $f$? – Siminore Sep 27 '12 at 13:30
As Siminore implies, this is wrong for arbitrary $f$ (take, for example, the step function: $f(x) = 0$ for $x \le 0$, and $f(x) = 1$ for $x > 0$). On the other hand, it's trivial for $f(x) = x$ (just choose $\epsilon = \delta$). So, which $f$ are you talking about? Edit: Actually, your condition is trivial. Choose any numbers $x_1, x_2$. The condition $0 < |x-a|<\delta$ doesn't mention either, so we can disregard it here. Now, choose $\epsilon = 2 |f(x_1) - f(x_2)|$, et voila... – Johannes Kloos Sep 27 '12 at 13:32
I am just being asked to solve a theorem. They idea comes from if the limit of a function, lim┬(x→a)⁡f(x)=L, exists, and x1 and x2 are extremely close to a, both f(x1) and f(x2) must be "approximately" L. That is, |f(x1)-f(x2)| – user42864 Sep 27 '12 at 13:44
Hint: You can make $|f(x_1) - f(a)|$ and $|f(x_2) - f(a)|$ small. – Serkan Sep 27 '12 at 13:50

closed as not a real question by Ayman Hourieh, Jasper Loy, BenjaLim, Thomas, MJD Sep 28 '12 at 1:55

It's difficult to tell what is being asked here. This question is ambiguous, vague, incomplete, overly broad, or rhetorical and cannot be reasonably answered in its current form. For help clarifying this question so that it can be reopened, see the FAQ.

1 Answer

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It's not true. Let $$ f(x) = \begin{cases} 0 & x \leq 0 \\ 1 & x > 0 \end{cases}$$ Let $\varepsilon \in (0,1)$ and let $a = 0$, $x_1 = \frac1n$, $x_2 = - \frac1n$. Then $|x - a| \leq \frac1n$ can be made arbitrarily small yet $|f(x_1) - f(x_2)| = 1$.

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