# For $\alpha > 0$, define the function: $g(x) = |x|^\alpha \cos(1/x^2)$, for $x \neq 0$, $g(x) = 0$, when $x = 0$

Define $$g(x) = \begin{cases}|x|^\alpha\cos(1/x^2) & x\neq 0\\ 0 & x = 0\end{cases}.$$

Determine what values of $\alpha>0$ is $g(x)$ differentiable at $x = 0$.

I had this question on an assignment and I got it wrong. Apparently the answer is $\alpha > 1$, however, I am having trouble understanding this, can someone help me?

In approaching this question, I determined that first I need continuity at $x = 0$. Therefore, I need a value for $\alpha$, such that $$\lim_{x \to 0} |x|^\alpha\cos(1/x^2) = 0.$$

So for this requirement, I need $\alpha\geq 1$. Is that correct? Because if I just test $(0.001)^{0.001}$, this gives something like $0.99$.

I will also need to ensure that $g'(x)$ from the left and right to be equal, so that there are no corners, or cusps at $x = 0$ (i.e. $g'(0) = 0$, since $\cos$ and $\sin$ oscillate).

Since $\cos(1/x^2)$ and $\sin(1/x^2)$ oscillate between $-1$ and $1$, I will need $2|x|^{\alpha - 3} \to 0$, as $x \to 0$, in order to have both the left and right derivatives the same.

So for the $(\alpha - 3)$, I would need this to be $\geq 1$. Therefore $\alpha > 4$.

Can someone help me to understand what I am getting wrong here?

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Where are you getting $\alpha-3$ from, and why would you need the exponent of $|x|$ to be $\ge1$ for it to $\to0$ as $x\to0$? – anon May 31 '12 at 0:48

A start: There is no continuity problem at $0$, for any $\alpha>0$. For fix $\alpha$. Note that $|\cos(1/x^2)|\le 1$, so for all $x$ we have $|g(x)|\le |x|^\alpha$. But for fixed $\alpha$, we can make $|x|^\alpha$ arbitrarily close to $0$ by choosing $|x|$ small enough.

So continuity is not the issue. For differentiability, there is no problem possibly at $x=0$. It may be best to go back to the definition of the derivative. So we want to know whether $$\lim_{h\to 0} \frac{|h|^\alpha\cos(1/h^2) -0}{h}$$ exists.

You will find that the limit does exist when $\alpha >1$, and doesn't when $0 \lt \alpha\le 1$. Since this is homework, and probably you can now continue, I will pause at this point.

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Hi Andre, thank you for your help. I actually already got my results from this assignment. The answer was given to be alpha > 1. What I don't understand is this though: if alpha is merely > 1, then for |h|^ (alpha - 1), if alpha is 1.00001, as h -> 0, we may have |0.00001|^(0.00001), and this goes to 1, not 0? – JackReacher May 31 '12 at 1:18
I guess I did not emphasize fixed $\alpha$ sufficiently in my answer, though I did use the word twice. OK, $\alpha$ is a fixed number, a constant. If it happens to be $1.00001$, that will not change. And if you pick $|x|$ to be small enough, like $10^{-444444444}$, then $|x|^{0.00001}$ will be close to $0$. – André Nicolas May 31 '12 at 1:26
Thank you. I understand now. Where I went wrong was that I made a wrong conclusion by observing a value for |x| that was not small enough. Thank you for helping me to understand this. – JackReacher May 31 '12 at 1:55
Yes, so why not use the definition? divide by $h$, being careful about sign, we get the term $|h|^{\alpha-1}$ if $h$ is positive, $-|h|^{\alpha-1}$ if $h<0$. Limit is $0$ if $\alpha>1$. Need to also show that things go bad if $0 \lt \alpha \le 1$. – André Nicolas May 31 '12 at 3:44
For $\alpha \le 1$, first $\alpha\lt 1$. Then absolute value of quotient is large when $1/x^2$ is a multiple of $\pi$. For $\alpha=1$, absolute value of quotient is $0$ is $1/x^2$ is a multiple of $\pi/2$, and $1$ if $1/x^2$ is a multiple of $\pi$. Each can happen with $x$ arbitrarily close to $0$. – André Nicolas May 31 '12 at 4:18

Any even function is differentiable at $0$ if and only if its right derivative there is $0$. Can you see why this is true? Try to prove it. Observe your function is an even function. Now the right derivative is

$$\lim_{x\to0^+}\frac{|x|^\alpha\cos(1/x^2)}{x}=\lim_{x\to0^+}x^{\alpha-1}\cos(1/x^2)=\cdots$$

because $|x|=x$ for $x\ge0$. Can you finish?

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