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I'm given with the following function:

$f(x) = x^n \arctan\left(\dfrac{1}{x}\right) , x\ne0 $ , and $f(0)=0$ .

I'm asked to:

  1. determine when is it continuous in $x=0$
  2. determine when is it differentiable in $x=0$
  3. determine when is it continuously differentiable in $x=0$

I have a problem with proving everything:

  1. When $n>0$ , we obviously have continuity because of the Squeeze Theorem. But how can I prove that when $n<0$ I don't have continuity ?

  2. When $n>0$, I don't have differentiability at $n=1$ . But how can I prove that in general , for $0<n<1$, the function isn't differentiable? and for $n>1$ it is?

  3. I guess that if I'll understand $1$ and $2$ I'll be able to understand $3 $.

Hope someone will help me

Thanks !

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I will give you some hints to solve this:

  1. Note that since $\lim\limits_{x\to \infty} \mathrm{arctan}(x)= \frac{\pi}{2}$, then $\lim\limits_{x\to 0^+} \mathrm{arctan}(\frac{1}{x})= \frac{\pi}{2}$. We know that for all $n<0$, the limit $\lim\limits_{x\to 0^+} x^n$ doesn't exist. What does this tell you about the limit $\lim\limits_{x\to 0^+} x^n\mathrm{arctan}(\frac{1}{x})$? Note that $\frac{x^n\mathrm{arctan}(\frac{1}{x})}{\mathrm{arctan}(\frac{1}{x})}=x^n$ For $n=0$, we have that $\lim\limits_{x\to 0^+} \mathrm{arctan}(\frac{1}{x})\neq \lim\limits_{x\to 0^-} \mathrm{arctan}(\frac{1}{x})$.

  2. If $0<n<1$, we can use the definition of derivative and the previous argument to show that the limit $\lim\limits_{x\to 0^+} f(x)/x=0$ doesn't exists.

  3. If $n>1$, we have $$ f'(x)=nx^{n-1}\mathrm{arctan}(\frac{1}{x})-\frac{x^{n}}{1+x^2} $$ Again, using 1 you should be able to say for which $n$ this function is continuos or not.

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For 1. you should be able to compute the limit of $f(x)$ for $x\to 0^+$ whatever is $n$. If the limit is different from $f(0)$ then the function is not continuous.

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