For what values is this function continuous

[lmstaples]

Homework Statement

F(x) \ = \ (x^{2})^{\frac{\alpha}{2}} \left[ \frac{\alpha sin \left( \frac{1}{x} \right)}{x} \ - \ \frac{cos \left( \frac{1}{x} \right)}{x^{2}} \right] \ if \ x \neq 0

F(x) \ = \ 0 \ if \ x \ = \ 0

Question basically says, for what values of alpha is this function continuous

Homework Equations

The Attempt at a Solution

My mind is torn between either: α = 0, for all α or some constriction on α being EVEN or ODD


thanks for any help

 

[Karnage1993]

You need the $\displaystyle \lim_{x \to 0} F(x) = F(0) = 0$ in order for the function to be continuous.

 

[lmstaples]

How do you do that for functions which have "two parts" to them?

Also, as x starts getting close to 0 everything is going to go crazy because of all the dividing by x's.

I remember doing a question a while back which was basically "is f(x) = sin(1/x) continuous at 0 if f(0) = 0?" I believe the answer was no because of sequential continuity?

 

[haruspex]

First, concentrate on the two terms inside the bracket. Can you see a reason to say that one of them makes the other irrelevant?
Next, assuming that a^b is defined to mean the positive real value when a > 0 and b > 0, can you see a way to simplify $\left(x^2\right)^{\frac{\alpha}{2}}$ that's valid for α > 0 and all x?

 

[lmstaples]

The main difference between the two terms in the brackets are:
1) the use of α in the sin term
2) the cos term is over a higher power of x

Possibly the cos term is irrelevant due to it not depending on α which is what in looking for, for continuity.

Or possibly the sin term is irrelevant as it tends to ∞ slower than the cos term?

The reason as two why it is (x^2)^(α/2) is because it was originally |x|^α

 

[haruspex]

The main difference between the two terms in the brackets are:
1) the use of α in the sin term
2) the cos term is over a higher power of x

What do you get if you collect them into a single fraction? Compare the two terms in the numerator. Can you see that one of them becomes irrelevant when compared to the other as x tends to 0?

The reason as two why it is (x^2)^(α/2) is because it was originally |x|^α

I feel the |x|^α form will be the easier to work with.

 

[lmstaples]

Sorry in advance for lack of latex, on a mobile.

Combining brackets gives us:

[α(x^2)sin(1/x) - xcos(1/x)]/[x^3]

I can't see how one term is particularly irrelevant.

As for |x|^α: by all means use that notation, I only changed it as that particular function [F(x)] actually came from finding the derivative of another function and I found it easier using the other notation.

If you could explain why that notation would be easier to work with would be helpful :)

 

[lmstaples]

Looking back, as x tends to 0 the cos term must become irrelevant.
e.g. If x = 0.000001,
x >> x^2

 

[haruspex]

[α(x²)sin(1/x) - xcos(1/x)]/[x³]

It's a little simpler than that: [αxsin(1/x) - cos(1/x)]/x²
What does each of the terms in the numerator do as x tends to 0?

 

[lmstaples]

Well the sin term will tend to 0 but the cos term will oscillate like crazy

 

[haruspex]

Well the sin term will tend to 0 but the cos term will oscillate like crazy

Right, so you can ignore the sin term. And of course x² = |x|². So what do can you reduce the whole expression to?

 

[lmstaples]

I believe I got it to reduce to [(x^2)^(α/2 - 1)][-cos(1/x)]

 

[haruspex]

I believe I got it to reduce to [(x^2)^(α/2 - 1)][-cos(1/x)]

Right, but I still think you're better off with the form [|x|^α-2][-cos(1/x)].
You've already described what -cos(1/x) will do as x tends to 0. So what do you need |x|^α-2 for the whole to converge to 0?

 

[lmstaples]

Would it not converge to 0 for any α >= 2