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For which values of parameter $a$ and $c$ function

$$ f(x)= \left\{ \begin{array}{ll} |x|^a\sin|x|^{-c} & \textrm{for $x \neq 0$}\\ 0 & \textrm{for $x=0$} \end{array} \right. $$

a) is continuous on the interval $[-1,1]$

b) it is differentiable in $[-1,1]$

c) the derivative is limited

this is my homework. i must to calculate $\lim_{x \to 0^-} $and $\lim_{x\to 0^+}$ this is the same lim.

So $\lim_{x \to 0^-}|x|^a\sin|x|^{-c}$

I don't know how calculate this... and what next b) and c) ..

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  • $\begingroup$ Is it $\sin \left( \left| x \right|^{-c} \right)$ or $\left( \sin \left| x \right| \right)^{-c}$? $\endgroup$ Mar 27, 2019 at 12:46
  • $\begingroup$ there are no parentheses in the notebook $\endgroup$
    – mona1lisa
    Mar 27, 2019 at 17:38

1 Answer 1

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I assume the functions is $$f(x)=|x|^a\sin(|x|^{-c})$$over $\Bbb R-\{0\}$. Note that this function is even, therefore$$\lim_{x\to 0}f(x)=\lim_{x\to 0^+}f(x)$$Now, we consider $3$ cases (in al of the cases we consider $x\ne 0$):

Case 1: $a>0$

The function is continuous by Sqeeze theorem since $$-|x|^a<f(x)<|x|^a$$

Case 2: $a=0$

The function is continuous only if $c<0$.

Case 3: $a<0$

In this case, the function has no limit at $x=0$ when $c=0$ or $c>0$. For $c<0$ we can write:$$\lim_{x\to 0}f(x){=\lim_{x\to 0}|x|^{a}\sin (|x|^{-c})\\=\lim_{x\to 0}\frac{\sin (|x|^{-c})}{|x|^{-a}}\\=\lim_{x\to 0}\frac{\sin (|x|^{-c})}{|x|^{-c}}\cdot{1\over |x|^{-a+c}}\\=\lim_{x\to 0}{|x|^{a-c}}}$$which is equal to zero only if $a>c$.

Conclusion

The values of $a,c$ for which $f(x)$ is continuous is as follows:$$a>0\\a=0,c<0\\c<a<0$$

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