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

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) ..

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

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

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