Epsilon-Delta limits, *not* for a straight line. Parabola example

[TupoyVolk]

So the general tactic for straight lines:

f(x) = 2x
Show by epsilon-delta definition of limit, as lim x->2, f(x) tends to 4.
let, ε>0, and choose 0<δ<ε/2

(|x - 2|<δ)→ |f(x) - 4| = |2x - 4| = 2|x - 2| < 2δ < ε
No problem, but what about for a parabola?

g(x) = ax² for some a in R.
Show as lim x -> b, g(x) tends to ab²

let, ε>0, and choose 0<δ<?
|x - 2|<δ
|g(x) - ab²| = |ax² - ab²| = a|x²-b²| ??
I don't have a clear idea what to do here; I tried;
a|(x+b)(x-b)| = a|(x+b)||(x-b)| < a|(x+b)|δ
Is it ok to simply choose:
0<δ<ε/|a(x+b)|
after all, it's just a positive number.

I really hope this post's question can be answered in one word beginning with y.

 

[TupoyVolk]

durr

 

[Hurkyl]

Is it ok to simply choose:
0<δ<ε/|a(x+b)|
after all, it's just a positive number.

I really hope this post's question can be answered in one word beginning with y.

In case others see this question, the answer is "no".

 

[TupoyVolk]

eh... why?

δ = ε|(x-a)/(f(x)-L)| always returns the desired ε interval, though.

It's the exact same as saying when f(x) = L ± ε, what is x.
Where's the problem with this?

It makes sense that δ should depend on x as the graph is not straight

Here is an example choosing ε = 1, for f(x) = x²

 

[wisvuze]

almost, you don't really know what "x" is, so that isn't really a fixed value for delta. You can get around this though, by thinking of upper bounds for x

 

[TupoyVolk]

How can you have a fixed delta if the epsilon is fixed and the function is a curve? You could easily choose the smaller delta if you wanted it to be 'within' the ε.

 

[wisvuze]

the delta depends on the epsilon, so it can change for different epsilons, that is true; but the "x" in | x - a | is any x within(a - delta, a + delta); for one, the x's are picked after the delta, so things are already bad if you have to depend on what "x" is.

 

[wisvuze]

eh... why?

δ = ε|(x-a)/(f(x)-L)| always returns the desired ε interval, though.

It's the exact same as saying when f(x) = L ± ε, what is x.
Where's the problem with this?

It makes sense that δ should depend on x as the graph is not straight

Here is an example choosing ε = 1, for f(x) = x²

your picture here is demonstrating that delta must depend on "a" as well as epsilon. This is correct, since x^2 is not uniformly continuous on R ( you can look up what this means if you don't know what it means already )
it seems to me that you are mixing up the "x" and the "a", but I'm not sure

 

[TupoyVolk]

I see, thank you!

So what's the trick to get a fixed delta for an f(x) = xⁿ

Upper bounds of x; ... take what would be "the smaller" δ (ie for x = (L + ε)^0.5) :/?

Excuse my niavity. It is not on the current course, I'm just 'annoyed' to only know how it works for straight lines.

Edit: Looked up, I understand. So, what to do for this case? If it is incredibly detailed, you don't have to waste your time.

 

[TupoyVolk]

So would it suffice to choose this delta for f(x) = x²;

δ = ε|(f^-1(L+ε)-a)/((L+ε)-L)|
δ = ε|(f^-1(L+ε)-a)/ε|
δ = |(f^-1(L+ε)-a)|

δ = |(L+ε)^0.5-a)|

ie: choose it for the highest value x can be so that such that |f(x)-L|<ε

Is that an OK 'choice' or does it have to be more specific?

∀x: 0<|x-a|<δ -> |f(x)-L| < ε is true here, but, the lowest value of x is not equal to f^-1(L-ε), I don't think that should matter, but for verification... is it a good value for δ?

 

[wisvuze]

I'll work out the x^2 example:

you want to show that for any epsilon ( e ) > 0 you can find a delta ( d ) > 0 such that for all x satisfying 0 < | x - a | < d , we get | x^2 - a^2 |< epsilon. ( we are assuming that we know that a^2 is the limit of x^2 as x -> a ).

we want to find a bound of | x - a | so that | x^2 - a^2 | < e is always ensured.
|x^2 - a^2 | = |x+a| |x-a|. back in the line example, we saw that similar computations showed us that if, for example, we wanted to ensure 5|x-a| < e, we could just take delta to be epsilon/5. So it seems like we should be able to take e/|x+a| as delta and be done with it, but as pointed out before, it doesn't work like this. However, if we could find an upper bound (which is we will find to be a constant ) to what |x+a| can ever be, then we would see this:
|x+a| | x- a| < |upperbound| | x-a |. Then picking any e, we can ensure |upperbound| | x-a| < e by setting delta to be e/|upperbound|, similar to the line example. But, once delta = e/|upperbound|, we see that |x^2 - a^2| = |x-a||x+a| < |upperbound| | x-a| < e and we should be done, since for any e, we have found a way (a delta ) to ensure |x^2 - a^2| < e.

So all that's required to do now is to find an upper bound for |x+a|.

Suppose we consider all x such that | x - a | < 1. I could have considered any constant c, ( for example, 1 , 5 , 10 .. or whatever ) . Then, if all x satisfying | x - a | < 1 also satisfies | x^2 - a^2 | < e then we are done, since we could just pick delta to be 1 ( or whatever c I picked ) . But we are not always this lucky, it might be the case that not all x satisfying |x-a| < 1 also satisfies |x^2 - a^2 | < e. So it might be the case that we will need a delta smaller than 1 ( note that if we could find a delta bigger than 1, then 1 certainly works too, since ( a -1 , a+1) would be a sub interval of (a- d , a + d ) ). But, things are still looking good, since we can now easily find an upper bound for | x + a|. We know that |x -a | < 1, so | x | - | a | <= |x - a | < 1 and we see that |x| < 1 + | a | . Then, |x + a| <= |x | + |a| < ( 1 + |a | ) + | a | = 1 + 2|a |. ( These follow from the triangle inequalities, you don't need to use them, but it is neater )
So now we have found an upper bound for | x + a | , we have that | x + a | < 1 + 2|a |. So as described before, we can now find our appropriate delta, since:

|x + a | | x- a | < (1 + 2|a |)|x-a| , so for any x so that | x - a| < e/(1 + 2|a |), we can ensure that (1 + 2|a |)|x-a| < e,
in particular, |x + a | | x- a | < (1 + 2|a |)|x-a| < e , ensuring that |x + a | | x- a | = |x^2 - a^2 | < e.
But, remembering that we initially ensured ourselves that | x - a| < 1, we of course must always have | x - a | < 1 hold. So we pick delta = min ( 1, e/(1 + 2|a |) ) , so that for any epsilon e > 0 | x - a | < min ( 1, e/(1 + 2|a |) ) implies | x^2 - a^2 | < e.

Picking the minimum between 1 and e/(1 + 2|a |) ensures that both |x - a| < 1 and |x -a | < e/(1 + 2|a |) are always held.

 

[TupoyVolk]

Perfect. I am incredibly satisfied, really.

Thank you for sharing these ideas with me!