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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 $x\to0$?2012-05-31

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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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    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 $0$.2012-05-31
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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?