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Given the function $ f(x) = x^a\sin(\frac{1}{x^b})$ if $x \ne 0,$ $0$ when $x=0$ for which values of $a,b$ is $f'(x)$ continuous at $0$?

So the derivative is $f'(x)=ax^{a-1}\sin(\frac{1}{x^b})-bx^{a-b-1}\cos(\frac{1}{x^b})$.

Is this continuous for all $a,b \in \mathbb{R}$ at $x=0$? Since this is one of those oscillating functions, or because $f(x)$ is differentiable at $x=0$ only for $a > 2$, is that not true?

Tianlalu
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clovis
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1 Answers1

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So, I used the definition of differentiability to see where the limit $f'(0)$ exists, and is therefore continuous.

for $f'(0) = lim_{x->0} \ ax^{a-1}sin(\frac{1}{x^b})=0$ when $a-1 > 0, $ so $a>1$. b should be able to be any real number.

and $f'(0) = lim_{x->0} \ bx^{a-b-1}cos(\frac{1}{x^b})=0$ when $a-b-1 >0$ and doesn't exist otherwise.

clovis
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