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Using the definition, prove that $\lim\limits_{x \to 10} 5 = 5$

Solution:

when I apply the definition, i get this

$0< |x - 10| < \delta \Rightarrow |5 - 5 | < \epsilon \Rightarrow 0 < \epsilon$

$0 < \epsilon \Rightarrow |x - 10| < \epsilon$ ,and $|x - 10| < \delta$

So i can take $\delta = \epsilon$, or less

Than

$\forall \epsilon, \epsilon >0, \exists \delta = \epsilon; \forall x \in D_f: 0< |x - 10| < \delta \Rightarrow |5 - 5 | < \epsilon $

Is it correct?

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    $\begingroup$ In fact, with $\textit{any}\;\delta > 0$ it works. $\endgroup$ – Ángel Mario Gallegos Jun 4 '15 at 19:07
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I think it's correct, but it's somewhat confusingly written. Let's see if we can clean it up:

Given $\varepsilon > 0$, you want to choose $\delta > 0$ so that if $|x-10|<\delta$ then $|f(x)-f(10)|<\varepsilon$. For this in fact any $\delta$ will work, since $f$ doesn't depend on $x$. So given whatever $\delta$ you like, $|x-10|<\delta$ guarantees $|f(x)-f(10)|=|5-5|=0<\varepsilon$.

I encourage you to use human language in your proofs of things like this. It's just confusing when written entirely in symbols.

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    $\begingroup$ I'm glad to see you made a comment about using symbols in writing a proof. I feel like symbols no more belong in a formal proof than "lol" or "g2g" belong in an English paper. $\endgroup$ – graydad Jun 4 '15 at 19:13
  • $\begingroup$ ok, Thanks a lot! $\endgroup$ – Gustavo Mega Jun 4 '15 at 19:15

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