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i'm trying to follow this proof of continuity of the function $f(x)=3x+4$ using Delta epsilon.

https://www.youtube.com/watch?v=WzOl_MwATf4

On minute 5:29, written in green, an inequality suddenly changes to an equality and i'm not sure why.

It would seem that for the definition to hold, you would need $$|f(y)-f(x)|<\epsilon$$ not $$|f(y)-f(x)|=\epsilon$$

And even if that was valid i'm not sure how he got there.

Is this correct? if so why?

enter image description here

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  • $\begingroup$ From the picture, it looks like the author fixed some arbitrary $\varepsilon>0$ and defined $\delta := \frac{\varepsilon}{3}$. Hence the equality $\endgroup$ – joeb Mar 14 '17 at 1:17
  • $\begingroup$ Hi, that's what he did, but i'm not sure why that leads to an equality, maybe i'm missing some concept in the definition? $\endgroup$ – Joaquin Brandan Mar 14 '17 at 1:19
  • $\begingroup$ That $3\cdot\frac{\epsilon}{3}$ refers to $3\delta$. It does not mean that $<$ is replaced by $=$. $\endgroup$ – Juniven Mar 14 '17 at 1:19
  • $\begingroup$ oh i dont mean that, i mean on the red arrows, the first one is an inequality, after that the author defines delta as $\epsilon / 3$, after doing that he replases delta with $\epsilon / 3$ on the second red arrow, and there he changes the inequality from the first red arrow into an equality. $\endgroup$ – Joaquin Brandan Mar 14 '17 at 1:21
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What you are seeing is a long chain of relations which would be read left-to-right if space permitted: $$ |f(y) - f(x)| = |(3y+4)-(3x+4)| = |3y-3x| = 3 |y-x| < 3 \delta = 3\cdot \frac{\epsilon}{3} = \epsilon $$ Each relation relates only the two expressions on either side. Since $\delta = \frac{\epsilon}{3}$, we know that $3 \delta = 3\cdot \frac{\epsilon}{3}$ at the second-to-last step.

But as they say, “a chain is only as strong as its weakest link.” So ignoring the intermediate steps by transitivity, we have $$ |f(y) - f(x)| < \epsilon $$ and that's exactly what was to be shown.

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    $\begingroup$ hi, thanks, why it it not written $|f(y)−f(x)|=|(3y+4)−(3x+4)|=|3y−3x|=|3(y−x)|<3 \delta <3⋅ \epsilon 3< \epsilon $ $\endgroup$ – Joaquin Brandan Mar 14 '17 at 1:24
  • $\begingroup$ ignore the minus epsilon at the last part, i cant remove it for some reason, i meant just epsilon $\endgroup$ – Joaquin Brandan Mar 14 '17 at 1:25
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    $\begingroup$ @JoaquinBrandan Because $\delta$ is not less than $\frac{\epsilon}{3}$; it's equal to it. And that's an em-dash (–) at the end which separates your comment from your signature, not an extraneous minus sign ($-$). $\endgroup$ – Matthew Leingang Mar 14 '17 at 1:25
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    $\begingroup$ aha! got it!. Would this be correct then? again ignore the minus sign at the end. $|f(y)−f(x)|=|(3y+4)−(3x+4)|=|3y−3x|=|3(y−x)|<3 \delta $ where $3\delta = 3⋅ \epsilon /3 = \epsilon$. $\endgroup$ – Joaquin Brandan Mar 14 '17 at 1:27
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    $\begingroup$ @JoaquinBrandan Yes, you got it. This is a common hurdle for students to overcome. Your question (and the followup) hit it on the head. $\endgroup$ – Matthew Leingang Mar 14 '17 at 1:30

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