Precise definition of a limit: Why is the delta chosen to be 1 and not √2 - 1

[Integrate]

We have the following limit to prove using delta-epsilon

[math]\lim_{x \to 1} \frac{x+1}{1+\sqrt{x}}[/math]
It can be manipulated into

[math]x-1| |\sqrt{x}-1||\frac{1}{{(1+\sqrt{x})}^{2}}|<\varepsilon[/math]
if our next step is to replicate
[math]\sqrt{x}-1[/math] in

[math]-1<|x-1|<1[/math] should it not be

[math]0<|\sqrt{x}-1|<\sqrt{2}-1[/math]instead of the given answer of [math]0<|\sqrt{x}-1|<1[/math] ?


[Problem and given solution here.][1]


[1]: https://www.math.ucdavis.edu/~kouba/CalcOneDIRECTORY/preclimsoldirectory/PrecLimSol.html#SOLUTION 12

 

[Dr.Peterson]

We have the following limit to prove using delta-epsilon

[math]\lim_{x \to 1} \frac{x+1}{1+\sqrt{x}}[/math]

Do you mean "Prove [imath]\lim_{x \to 1} \frac{x+3}{1+\sqrt{x}}=2[/imath]" as in the source, or are you doing a different problem?

if our next step is to replicate [imath]\sqrt{x}-1[/imath] in [imath]-1<|x-1|<1[/imath] should it not be [imath]0<|\sqrt{x}-1|<\sqrt{2}-1[/imath]
instead of the given answer of [imath]0<|\sqrt{x}-1|<1[/imath] ?

I'm not sure what you mean by "replicate".

You're apparently referring to this sentence:

​

If [imath]0<x<2[/imath], then [imath]0<\sqrt{x}<\sqrt{2}[/imath], so [imath]-1<\sqrt{x}-1<\sqrt{2}-1<1[/imath], so they are right that [imath]0\le|\sqrt{x}-1|<1[/imath].

Your conclusion, that [imath]0\le|\sqrt{x}-1|<\sqrt{2}-1[/imath], is wrong, as it is not true if, say, [imath]x=0.01[/imath] so that [imath]|\sqrt{x}-1|=0.9>0.414\approx\sqrt{2}-1[/imath].

You overlooked the fact that in turning an inequality into an absolute value inequality, you need to consider both positive and negative values. What we are doing here is expanding the interval on one side to balance it.

 

[Integrate]

Do you mean "Prove [imath]\lim_{x \to 1} \frac{x+3}{1+\sqrt{x}}=2[/imath]" as in the source, or are you doing a different problem?

I'm not sure what you mean by "replicate".

You're apparently referring to this sentence:

View attachment 30242​

If [imath]0<x<2[/imath], then [imath]0<\sqrt{x}<\sqrt{2}[/imath], so [imath]-1<\sqrt{x}-1<\sqrt{2}-1<1[/imath], so they are right that [imath]0\le|\sqrt{x}-1|<1[/imath].

Your conclusion, that [imath]0\le|\sqrt{x}-1|<\sqrt{2}-1[/imath], is wrong, as it is not true if, say, [imath]x=0.01[/imath] so that [imath]|\sqrt{x}-1|=0.9>0.414\approx\sqrt{2}-1[/imath].

You overlooked the fact that in turning an inequality into an absolute value inequality, you need to consider both positive and negative values. What we are doing here is expanding the interval on one side to balance it.

ahhhh you are right. I am just to rote sometimes. So essentially we had to move up our upper bound until it made sense. In this case it was up to one.

 

[Dr.Peterson]

ahhhh you are right. I am just to rote sometimes. So essentially we had to move up our upper bound until it made sense. In this case it was up to one.

RIght. Or you could say we have an off-center interval, and have to find a symmetrical interval that contains it, which amounts to taking the larger of the two distances from the center.

This is very common in epsilon-delta proofs.

 

[Steven G]

Although not incorrect, you should not have written -1< |x-1|<1. Actually you probably meant to write |x-1|<1 or -1<x-1<1.
Then 0<x<2
So sqrt(x) < sqrt(2)

 

[Integrate]

RIght. Or you could say we have an off-center interval, and have to find a symmetrical interval that contains it, which amounts to taking the larger of the two distances from the center.

This is very common in epsilon-delta proofs.

Where might I find more of examples of these to practice? I am using the james stewart calc book and haven't come across them in the problem exercises.

 

[Dr.Peterson]

Where might I find more of examples of these to practice? I am using the james stewart calc book and haven't come across them in the problem exercises.

Maybe it's not as common as I was imagining! In my mind, it's almost always involved (in principle) because the actual points on the graph that are delta away in either direction correspond to an imbalanced interval vertically, and vice versa. But often, as in the several quadratic examples in your source, that is handled indirectly by taking a bound on one factor, so you don't notice it.

Since in an ordinary calculus course we don't really need to do many of these proofs (using properties of limits instead), I don't think it's too important to have a lot of practice with the harder cases. (Which could be why you don't see it in your book?)

 

[Integrate]

Maybe it's not as common as I was imagining! In my mind, it's almost always involved (in principle) because the actual points on the graph that are delta away in either direction correspond to an imbalanced interval vertically, and vice versa. But often, as in the several quadratic examples in your source, that is handled indirectly by taking a bound on one factor, so you don't notice it.

Since in an ordinary calculus course we don't really need to do many of these proofs (using properties of limits instead), I don't think it's too important to have a lot of practice with the harder cases. (Which could be why you don't see it in your book?)

Thank you. I’ll move on to the next section then!