Prove Limit l of f(x) is Zero with abs(x) < ε

[Miike012]

Problem: Determine the limit l for the given a, and prove that it is the limit by showing how to find a δ such that abs(f(x) - l)< ε for all x satisfying 0< abs( x - a) < δ.

** abs = absolutue value of...

Solution:
f(x) = x(3-cos(x^2))

The limit l is zero. hence the product of some number times zero is zero.

abs(f(x) - 0)< ε = abs(x(3-cos(x^2)).

abs((3-cos(x^2))≤ 1, abs(x(3-cos(x^2))< abs(x)

Thus for every number x, If abs(x) < ε Then abs(x(3-cos(x^2)),

Which finishes the proof, concluding that the limit of f(x) near zero is zero.

 

[gb7nash]

For future reference, use | for absolute value. It's easier to read.

f(x) = x(3-cos(x^2))

The limit l is zero.

Is the problem to show that $\lim_{x \to 0} f(x) = 0$?

hence the product of some number times zero is zero.

I'm not sure where this comes from.

abs(f(x) - 0)< ε = abs(x(3-cos(x^2)).

The bolded part is good start. I'm not sure why you're equating epsilon to |x(3-cos(x²)|.

abs((3-cos(x^2))≤ 1, abs(x(3-cos(x^2))< abs(x)

Neither of these statements are true. Do you see why?

Thus for every number x, If abs(x) < ε

It's unclear how you concluded the bolded part.

Then abs(x(3-cos(x^2))

Then what? Is there supposed to be an equation here?

 

[Miike012]

"I'm not sure why you're equating epsilon to |x(3-cos(x2)|."
Are you asking me why i said |x(3-cos(x2)| < ε?

"Neither of these statements are true. Do you see why?"
|x(3-cos(x2)| is not less than or equal to 1.. is that right?
abs(x(3-cos(x^2))< abs(x), I am not sure why that is not correct...?

"It's unclear how you concluded the bolded part."

Because I said |x(3-cos(x2)| < ε, then I found out |x(3-cos(x2)| < |x|
Then I assumed |x(3-cos(x2)| < |x - 0| < ε Thus |x| < ε

 

[gb7nash]

"I'm not sure why you're equating epsilon to |x(3-cos(x2)|."
Are you asking me why i said |x(3-cos(x2)| < ε?

No. I'm asking why you're setting ε = |x(3-cos(x^2)|.

"Neither of these statements are true. Do you see why?"
|x(3-cos(x2)| is not less than or equal to 1.. is that right?

Yes.

abs(x(3-cos(x^2))< abs(x), I am not sure why that is not correct...?

-1 ≤ cos(x²) ≤ 1, so:

|x(3-cos(x²)| ≥ |x(3-1)| = |2x| > |x|. This is the opposite of what you want.

 

[Miike012]

-1 ≤cos(x2) ≤ 1, I am assuming you picked +- 1 because of the range of the function?
In general when proving limits how do I determine the appropriate interval for my function? For example If I had some polynomial g could I say -n ≤g(x) ≤ n, where n is an integer?

 

[gb7nash]

-1 ≤cos(x2) ≤ 1, I am assuming you picked +- 1 because of the range of the function?

Correct. Cosine is bounded between -1 and +1.

In general when proving limits how do I determine the appropriate interval for my function? For example If I had some polynomial g could I say -n ≤g(x) ≤ n, where n is an integer?

You don't want to use integers. That's what the epsilon is for. By letting ε > 0, this covers every case.

 

[Miike012]

The books solution was...
-1 ≤ cos(x2) ≤ 1, so |(3-cos(x2)| <=4, and thus |x(3-cos(x2) - 0| = |x|*|x(3-cos(x)|<= 4|x|.

So we take delta = ε/4.

So in the first step they said -n ≤ cos(x2) ≤ n where n = 1, so If I had another situation like this example but the function was say a polynomial... then How would I pick n.
I hope I am making sense.

I understand that I have to use epsilon when I am saying |f(x) - l|.

 

[SammyS]

I know that gb7nash asked you this, but I didn't see any answer.

Precisely, what limit are you trying to evaluate? Mostly, what is a ?

$\displaystyle\lim_{x\,\to\,a}(x(3-\cos(x^2)))$

 

[Miike012]

a is zero, so the limit is zero..
Sorry I forgot to put the value of a in my problem.