Differentiability of $x^\alpha \sin(x^{-\beta})$ at $x = 0$

Question

\begin{align*} f(x) = \left\{\begin{array}{ll} 0 & \text{ if } x=0\\ x^\alpha \sin(x^{-\beta}) & \text{ otherwise } \end{array}\right. \end{align*}

Determine the values of $\alpha$ and $\beta$ for which this function is differentiable at $x=0$.

I found the derivative, but I don't know what to do after...

What did you find for the derivative, and how did you find it? By the way, are there conditions on $\alpha$ and $\beta$? — Jason Zimba, Mar 27 '14 at 02:23
Yes, the conditions on $\alpha$ and $\beta$ must be found. I differentiated the second function with respect to x — kiwifruit, Mar 27 '14 at 02:24
@kiwifruit I assume that you just used product and chain rule for the derivative? If so, that only works away from $0$. — , Mar 27 '14 at 02:27
By differentiating the function, you have presupposed what you were trying to prove. Or else you have found a result for $x>0$, where the differentiability is not in question. — Jason Zimba, Mar 27 '14 at 02:27
Yes, I found the derivative only for the second part using the product & chain rule. Not sure what to do with the 0 — kiwifruit, Mar 27 '14 at 02:28
@JasonZimba I see! What would the right approach be then to prove differentiability? — kiwifruit, Mar 27 '14 at 02:28

score 2 · Accepted Answer · answered Mar 27 '14 at 02:29

2

Hint: Remember the definition of the derivative at $0$:

$$f'(0) = \lim_{x \to 0} \frac{f(x) - f(0)}{x} = \lim_{x \to 0} x^{\alpha - 1} \sin\left(x^{-\beta}\right)$$

Now the sine function oscillates as the input goes to zero or to infinity, so the only way the limit can possibly exist is if the term $x^{\alpha - 1}$ kills the oscillation by tending to zero. This suggests that you might want to consider cases based on $\alpha > 1$, $\alpha = 1$ and $\alpha < 1$.

Can you take it from here?

answered Mar 27 '14 at 02:29

Well, it seems like only $\alpha >1$ would give the necessary result. But are there any conditions for $\beta$? – kiwifruit Mar 27 '14 at 02:31
@kiwifruit Consider the case $\alpha = 1$; a condition on $\beta$ is necessary there. – Mar 27 '14 at 02:32
So if $\alpha=1$, we get the oscillating sin. Then must we force it to be some specific value like 1? – kiwifruit Mar 27 '14 at 02:40
@kiwifruit Not quite. The question is whether $\lim_{x \to 0} \sin\left(x^{-\beta}\right)$ exists: Again, considering cases of $\beta$ will be helpful. – Mar 27 '14 at 02:43
It does not exist of $\beta=0$, but it will be some value for the other cases? – kiwifruit Mar 27 '14 at 02:44
@kiwifruit If $\beta = 0$, then you're considering $\lim_{x \to 0} \sin(1)$. – Mar 27 '14 at 02:45
So it would be a fixed value in this case, and sin(0) in all other cases? – kiwifruit Mar 27 '14 at 02:49

score 2 · Answer 2 · answered Mar 27 '14 at 02:51

Just to clarify some cases. I came too late $$\lim_{x \to 0}\frac{x^\alpha sin(x^{-\beta})-0}{x-0}=\lim_{x \to 0}x^{\alpha-1} sin(x^{-\beta})$$ Then is differentiable if the previous limit exists. We have the case $\alpha>1$, $\alpha=1$ $\alpha>1$ and the cases for $\beta=0$,$\beta>0$, $\beta<0$

$\alpha>1$ $\beta=0$ clearly converges to $0$

$\alpha<1$ $\beta=0$ clearly diverges

$\alpha=1$ $\beta=0$ clearly converges

$\alpha>1$ $\beta<0$ converges to $0$

$\alpha<1$ $\beta<0$ we have a limit of the form: $$\lim_{x \to 0}\frac{sin(x^{-\beta})}{x^{1-\alpha}}$$ Take $u=x^{-\beta}$ then $x^{1-\alpha}=u^{\frac{\alpha-1}{\beta}}$ $$\lim_{u \to 0}\frac{sin(u)}{u^{\frac{\alpha-1}{\beta}}}$$ Here if $\frac{\alpha-1}{\beta}<1$ it converges and in another case diverges

$\alpha=1$ $\beta<0$ clearly converges

Differentiability of $x^\alpha \sin(x^{-\beta})$ at $x = 0$

2 Answers2