Why Pick Delta as 1 in Limits of Functions?

[semidevil]

I"m doing some problems regarding limits of functions, and using the epsilon delta defnition, I really have no idea on what they do, and how they pick the delta.

for example:

lim as x to 3 of | 5-3/x+3| = 1/3. after some siimplication, this turns to 4/3|x-3 / x+3| < epsilon. and to find a bound for |x+3| we set delta to 1. and this makes -1 <x-3 < 1. then, they add 6 to both sides, and this becomes -5 <|x+3|<7. after that, it is pretty straight forward.

first question, why did they pick delta as 1? and also, why did they add 6 to the left and right side?

and then, there are other problems such as as limit of x to 2 of 1/ 1 - x = - 1. they pick delta as 1/2, and what do they do from there?

 

[arildno]

I have absolutely no idea of what function you're talking about!
Which of these functions do you mean:
a) $$f(x)=|\frac{5-3}{x}+3|$$
b)$$f(x)=|5-\frac{3}{x}+3|$$
c)$$f(x)=|5-\frac{3}{x+3}|$$
Neither of these functions, I might add, have the limit value you are talking about at x=3

 

[dextercioby]

Arildno,i think it's
$$f(x)=|\frac{5-3}{x+3}|$$

Incidentally,this one converges to 1/3,when x----->3

Daniel.

 

[arildno]

"When we have eliminated the impossible, whatever remains, however improbable, is the truth"
(Or something like that..)

 

[mathwonk]

i think its (5-x)/(x+3) as x-->3.

and the answer is that quotient limits are hard. you soften them up by taking delta less than something easy like 1, and then go from there.

 

[matt grime]

The way to do practically all calculus questions splits into two parts.

Generic show lim x tends to a of f(x) is f(a)

1. Just suppose that |x-a|< d play around with |f(x)-f(a)| until you've got something bounding it above in terms of d.

eg.

to show x^2 is continuous on [0,1]

Pretend |x-a|< d , then |x^2-a^2| = |x-a||x+a| < |x+a|d <=2d since x and a both are in the interval [0,1], x+a is at most 2.


So, given e > 0 let d = e/2, then |x-a| <d implies by exactly the same reasoning that |x^2-a^2|<e, so x^2 is continuous at a.

2. Sometimes we don't get such a nice answer, so we can also assume d to be less than some number, for instance 1 if that helps.

For instance to show x^2 is continuous at all points of R, let us assume |x-a|<d<1, then by the above

|x^2-a^2| < d|x+a|

but -1<x-a<1, so 2a-1<x+a<2a+1, ie |x+a| < |2a+1|

so given e >0, let d be e/(2a+1).

The problem I find in students is that they actually think the lecturer plucked d out of thin air by a lucky guess that seemed to work, and can't figure out how they'd guess it. This owes to the fact that mathematicians like to make things look neat when presenting it so they don't "show their working".

 

[Galileo]

to show x^2 is continuous on [0,1]

Pretend |x-a|< d , then |x^2-a^2| = |x-a||x+a| < |x+a|d <=2d since x and a both are in the interval [0,1], x+a is at most 2.


So, given e > 0 let d = e/2, then |x-a| <d implies by exactly the same reasoning that |x^2-a^2|<e, so x^2 is continuous at a.

You really need to take cases, for when epsilon gets big, the result may not hold. |x+a| is not at most 2, because x doesn't have to lie in the interval [0,1]. x lies in the interval (a-d,a+d).

In your example, take for instance a=1, and e=10, then d=5,
but x=5 satisfies |x-1|<5, but |x^2-1^2|=|25-1|=24 >10.

|x+a| is at most 2, if you restrict x to the interval [0,1], this is not possible if a=1, because then d would have to be zero.
Better choose |x+a| at most 3, you can restrict x to the interval [0,2] by choosing d such that |x-a|<2-a. So choose d=min(e/3,2-a) to be on the safe side.

Hopefully I got that right. This is indeed what makes limits difficult.

 

[matt grime]

Erm, I think you'll find that x and a are in the interval [0,1] because that was part of the hypothesis. I was proving it continuous on the interval [0,1] (with the obvious subspace topology).

 

[Galileo]

Here an easier and more general way:

Let |x-a|<d.
Write |x+a|=|x-a+2a|<=|x-a|+2|a|.
Then |x^2-a^2|<=|x-a|(|x-a|+2|a|)<d^2+2|a|d

So we want d^2+2|a|d to be smaller then any e.
If d<1, then d^2+2|a|d<1+2|a|, so take:

$$\delta = min\{\frac{\epsilon}{1+2|a|},1\}$$

The argument holds for any $a\in \mathbb{R}$.

 

[Galileo]

Erm, I think you'll find that x and a are in the interval [0,1] because that was part of the hypothesis. I was proving it continuous on the interval [0,1] (with the obvious subspace topology).

Okay, I thought you meant continuous on a particular subset $D\subset \mathbb{R}$, with $D=[0,1]$.

 

[matt grime]

yes, that was the second argument i posted when i removed the condition that x lie in [0,1]