$f_n(x)=\sqrt[n]{1+x^n}$. First of all, I have to find its different limits so that I can find $f(x)$. This is where I have a problem: if $x> 1$, then $x\le f_n(x)=f(x)\le 2x$, but I can't get the precise approximation. If $x\le1$, then $\lim\limits_{n\to \infty}{\sqrt[n]{1}}=1\le \lim\limits_{n\to \infty}{f_n(x)}\le \lim\limits_{n\to \infty}{\sqrt[n]{2}}=1$.
What happens when $x$ is between 0 and -1, or when it is <-1?
I could use some help here.