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Hello,
I'm having a lot of trouble with my practice exam, below are two that are the hardest for me. I also attached an image of them in case the notation is difficult to understand.

Let f(x) = abs(x)/x
where x is not 0, and f(0) = 0. Prove that lim x->0 f(x) does not exist.

I'm not sure how to even set this up I know
I know L is arbitrary
Let ε>0 = something in terms of L
Let δ be arbitrary
And let x = something in terms of δ

I'm also having trouble with this one.

Let f(x) = 2x^2 -1 Prove directly from the definition that f(x) is continuous everywhere.
Then f(x) is continuous at a iff for every ε>0, ∃ δ>0 such that

|x−a|<δ --->|f(x)−f(a)|<ε

|f(x) - f(a)| <  
|2x^2 -1 - 2a^2 -1| = |2x^2 -2a^2|
= 2|x^2-a^2| <  
okay so |x^2-a^2| <  /2
I'm pretty sure I'm missing something and need |x^2-a^2| simplified more, I don't know what I can do though.
I could take out the (x+a)|x-a| but is δ allowed to have x and a terms in it? so |x-a|< /2(x + a) ?

Answer
Questioner:Rachel
Country:Maryland, United States
Category:Advanced Math
Private:No
Subject:Limit proofs
Question:

Hello,
I'm having a lot of trouble with my practice exam, below are two that are the hardest for me. I also attached an image of them in case the notation is difficult to understand.

Let f(x) = abs(x)/x
where x is not 0, and f(0) = 0. Prove that lim x->0 f(x) does not exist.

There are 10 important principles to use here:

1. Know the definitions of the terms.
2. Know the definitions of the terms.
3. Know the definitions of the terms.
4. Know the definitions of the terms.
5. Know the definitions of the terms.
6. Know the definitions of the terms.
7. Know the definitions of the terms.
8. Know the definitions of the terms.
9. Know the definitions of the terms.
10. Apply the definitions of the terms.

....................................
Part 1:  Easy -- show that  lim [x -> 0+] f(x)  and lim [x -> 0-] f(x) are not the same.
The definition of lim [x -> a] f(x) = L  is:  WHENEVER x is near a, f(x) is near L.
lim [x -> 0+] means 'from the right',

which means the right of x = 0,  

which means for positive values of x.

But when x is positive |x|, by definition, is equal to x.

Then f(x) = |x|/x = x/x = 1.  So lim f(x) = 1.

But (as you will show)  lim [x -> 0- ] f(x) = -1.

So there is no L, such that  WHENEVER x is near 0, f(x) is near L.  

NOTE:  This is how you will handle all those exercises coming up next week that look like this:

      |   (one thing),  x < some number
f(x) = |
      |  (another thing), x >= that number

and prove that  lim [x -> that number] exists (or does not exist)

...........................................................

I'm not sure how to even set this up I know
I know L is arbitrary
Let ε>0  =  something in terms of L
Let δ       be arbitrary
And let x = something in terms of δ

I'm also having trouble with this one.

Let f(x) = 2x^2 -1 Prove directly from the definition that f(x) is continuous everywhere.
Then f(x) is continuous at a iff for every ε>0, ∃ δ>0 such that

|x−a|<δ --->|f(x)−f(a)|<ε

|f(x) - f(a)| <  
|2x^2 -1 - 2a^2 -1| = |2x^2 -2a^2|
= 2|x^2-a^2| <  
okay so |x^2-a^2| <  /2
I'm pretty sure I'm missing something and need |x^2-a^2| simplified more, I don't know what I can do though.
I could take out the (x+a)|x-a| but is δ allowed to have x and a terms in it? so |x-a|< /2(x + a) ?  
..............................
WARNING: I WON'T BE ABLE TO MAKE THOSE SYMBOLS -- I JUST USE  e,d.
If you do an epsilon-delta proof, you try to show that given e, you can always find d.

Will  d depend on e?  Of course.
Will  d also depend on 'a'?  Probably.  If it does not, you have not just continuity, but something called uniform continuity.  Forget that -- it's for the pro's.

Why does d depend on a?  If the graph of f(x) around x = a is very steep, you need a smaller d.
................
So let's try.  First we apply those ten principles:

Prove that  lim [x-> a] f(x) = f(a), or: (as you noted)

Given e, we can find d such that

|f(x) - f(a)| < e,  whenever  | x - a | < d

|f(x) - f(a)| < e

|2x^2 - 1 - 2a^2 - 1| < e

|2x^2 - 2a^2| < e

| x^2 - a^2| < e/2

|x - a||x + a| < e/2

|x - a| < e/(2|x + a|)

OK, now we must get that 'x' out of the right side.  
We ask at this point, How small can the RHS be?  That's how small we must make  |x - a|.

A fraction gets small when the bottom gets big.

How big can x+a be?  We now make reasonable assumptions about a and d.  (THIS is the tricky part.)

1.  a is not zero.
2.  d is not to be more than 1.  (It's supposed to be small, right?)

Then   | x - a | < 1, so
- 1 < x - a < 1
a - 1 < x < a + 1
+ a      +a   +a
--------------------
2a - 1 < x+a < 2a + 1

Let's just assume a is positive, the arithmetic is easier.

So the largest |x+a| can be is  2a+1.
Then the smallest that  e/(2|x + a|) can be is  e/(2(2a + 1))

We are done.  Just pick  d < e/(2(2a + 1)) and we will always have

|f(x) - f(a)| < e

.... for positive a, anyway.  For negative a, the work is (slightly) different.

The reasoning is always a little tricky;  that's why I was so verbose in writing this out -- I wanted to make sure I was getting it right, too.

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