# When is the following function continuous? differentiable? …

I'm given with the following function:

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

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 !

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.
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.