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


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

  • $\begingroup$ 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

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.

  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ @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
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ @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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.