Question

Why is the following assertion not necessarily true: lim x->a f(x) = lim 2x->2a f(x)?

Answer 1

A counterexample will suffice.

Answer 2

Actually, this is true and I can prove it with epsilon-delta formality, if you are a college student who has seen it.

Answer 3

I am a college student, and I have seen epsilon-delta proofs. Here's the official problem in my book:
Give an example of a function f for which the following assertion is false: If |f(x)-l|<E when 0<|x-a|<d, then |f(x)-l<E/2 when 0<|x-a|<d/2. (Where E is epsilon and d is delta).

The first half seems to imply that lim x->a f(x)=l, and the second half (which I'm supposed to show is false) seems to imply that lim 2x->2a 2f(x)=2l; we factor out a 2 from each side, and we arrive at lim 2x->2a f(x)=l. Am I doing anything wrong?

Answer 4

Ah ... ok. The problem from your book is something else and it is true.

Answer 5

Oh... what did I do wrong?

Answer 6

by which I mean it is indeed the case that if .... < d => ... < e then you can't necessarily say that ... < d/2 => .... < e/2

Answer 7

Ah, okay. Any function you can think of that makes the statement false?

Answer 8

For example f(x) = x^(1/3) . Look at the limit as x --> 0

Answer 9

Now it is indeed true that for all e > 0 there exists a d such that

|x-0| < d => |x^(1/3) - 0|< e

You can see how intuitively this is going to fail because for x close to zero, the function approaches 0 slowly. So if you halved the distance to zero, i.e., looked at d/2, it would not necessarily be the case that
|x^(1/3)| < e

Answer 10

To formalize this idea, start as always with an e, let it be 0 < e < 1.
Then find explicitly the tightest value of d you can for which

|x-0| < d => |x^(1/3) - 0|< e

Then show it is false that

|x-0| < d/2 => |x^(1/3) - 0|< e/2

Answer 11

Yes?

Answer 12

Hm... one sec...

Answer 13

Oh, okay. Let me see...

Answer 14

I still don't really see how that works. What do you mean with the "tightest" value of d? The smallest value of d, so that we restrict x as much as possible?

Answer 15

Actually let's got back to f(x) = x^(1/3)

For e > 0 and e < 1, then for d = e^3

|x - 0| < d => |x^(1/3) - 0| < 0

Answer 16

< 0?

Answer 17

Where did that come from?

Answer 18

< e

But if
|x - 0|<d/2 = e^3/2, then it is NOT true that |x^(1/3)|<e/2 because

|x|<e^3 / 2 => |x^(1/3)| < e/2^(1/3) and 1/2^(1/3) > 1/2

Answer 19

There's some formalism here, but do you understand the idea? Draw the graph of f(x) = x^(1/3) over the domain [-1,1] and see how this function is behaving.

Answer 20

That all makes sense except for the last statement, I have no idea where that came from.

Answer 21

Just literally taking the cube root of both sides of |x|<e^3 / 2

Answer 22

Then we'd get x^(1/3)<e/2^(1/3)? I still don't see what you did

Answer 23

The point is
\[\frac{\epsilon}{2^{1/3}} > \frac{\epsilon}{2}\]

Answer 24

We can't get the sharpness or tightness of the inequality that we need.

To make it explicity, choose an epsilon. Let's take epsilon, e= 1/1000

Answer 25

then for d= 1/10
|x - 0| < d => |x^(1/3) - 0| < e

Answer 26

Sure.

Answer 27

But for d/2 = 1/20. It is the case that ....

Answer 28

1/8000 < e

Answer 29

Right?

Answer 30

Scuse me, 1/8000<e/2

Answer 31

Oh, my first qualm was right. Sorry. Take f(x) = x^2 instead. And let's take epsilon = 1/100. Then for delta, d = 1/10
|x - 0| < d => |x^2 - 0| < e

Answer 32

No. Stay with our first example: f(x) = x^1/3

Answer 33

Haha, ok.

Answer 34

So for f(x) = x^(1/3). Take e = 1/10. Then for d = 1/1000
|x-0| < d => |f(x)| < e

Answer 35

That's what got us caught up above

Answer 36

Now, for d/2 = 1/2000
|x| < d/2 => |f(x)| < 1/2000^(1/3) = 1/(2^(1/3).10)

Answer 37

What does 1/(2^(1/3).10) mean?

Answer 38

Never mind actually, I get that.

Answer 39

cube root of 2 times 10.

Now to show it is not true that
|x| < d/2 => |f(x)|<e/2

we need to show that there is an x such taht
|x| < d/2 and |f(x))| >= e/2

Answer 40

so we're looking for an x such that
|x| < 1/2000 and x^(1/3) >= 1/20

Answer 41

and of course there infinity many such x because that last inequality is equivalent to
x >= 1/8000

Answer 42

Ah.

Answer 43

You have no idea how confused I am right now but I'm going to go read over this and see if I can make sense out of it. Thanks.

Answer 44

Like I said above somewhere, draw the graph of f(x) = x^(1/3) and sketch out the delta/epsilon inequalities on the graph and you'll see why this works.

Answer 45

All right, that makes sense. But how did you know to take f(x)=x^(1/3) and what values to pick for e and d?

Answer 46

You can see that this would work for a linear function like f(x) = ax

Answer 47

So we needed a function that approaches something slower than linearly. Again, the graph gives you the picture and if we were in a room with a white board--vs. the blah drawing tool here--I'd show you.

Answer 48

Anyway ... sorry for a bit a confusion there.

Answer 49

Ahhh, ok. Thanks a lot.