Question on proofs of limits

Question

If I am trying to prove a limit say

$\lim(5x-3)=2$ as $x\to1$ then

Let $\epsilon$ be given. the aim is to find a $\delta>0$ such that whenever

$0<|x-1|<\delta$ then $|f(x)-2|<\epsilon$

My question is if I work from $|f(x)-2|<\epsilon$ to isolate $x$ an find a suitable $\delta>0$ then do I have to go back and show the implication or is the proof done?

In my book they did show the implication for this one because its nice and easy but for more complicated ones they just isolate $x$ and choose $\delta>0$ to be the shortest distance to one of the endpoints and then they just take $\delta>0$ to be this value and say the proof is done without actually showing the implication.

Thanks

Answer 1

The given function in this example is linear which makes life much easier to show the implications.

In general lt takes real analysis to show the implications but in order to have a valid proof the implications should be proved.

Reading some proofs in an advanced calculus textbook is useful to an inexperienced student.

Answer 2

Isolate $5$ here.

$|f(x)-2|<\epsilon\iff |5x-3-2|<\epsilon\iff5|x-1|<\epsilon\iff|x-1|<\epsilon/5$

Let $\delta = \epsilon/5$

So,

$|x-1|<\delta\iff|f(x)-2|<\epsilon$

Answer 3

It's a matter of whether or not the logic can be "reversed". If your argument looks like

1. $|f(x)-2| < \epsilon$
2. implies stuff
3. implies more stuff

$\qquad \vdots$

$1000000. \quad \text{implies} \quad 0 <|x-1| < \delta $

Then it does not logically follow that $0 <|x-1| < \delta \implies |f(x)-2| < \epsilon$

---

But if your argument looks like

1. $|f(x)-2| < \epsilon$
2. iff stuff
3. iff more stuff

$\qquad \vdots$

$1000000. \quad \text{iff} \quad 0 <|x-1| < \delta $

Then it does logically follow that $0 <|x-1| < \delta \implies |f(x)-2| < \epsilon$

---

Stylewise, the above proof is a form of laziness because it forces the reader to check that the logic really can be traversed in the opposite direction.

Studentwise, it's dangerous because you don't really know that the logic works in the other direction unless you work out that it can. So why not give your instructor a break and show him that it works out that way.

Answer 4

I would say even more: showing the implication $0<|x-1|<\delta$ then $|f(x)-2|<\epsilon$ is the only thing that constitutes a proof. The part where you manipulate equations in order to find a $\delta$ that will work is not even a necessary part of the proof. You might want to continue to write that part on your homework assignments and exams (where it shows that you knew how to develop the formula you needed for the proof and did not just make a lucky guess), but in actual professional mathematical papers that kind of thing is likely to be left out.

I suspect that the examples in your book were not intended to be read as proofs but rather were intended to show the kinds of work that one might do while preparing to start a proof. You may be expected to figure out on your own how the actual proof will go, once you have been shown a suitable formula for $\delta$ for the particular case in question. We would have to see a complete example in its exact original wording on order to way whether it was meant this way or meant as the proof itself.