The given function is :
$f(x) = \lim_{n\to\infty} \sqrt[n]{1+x^n}$ $ ;(x \ge 0)$
The function variable is $x$ and not $n$. This seems strange to me. All I can think of is that $f(x)$ is a family of curves which depends on the parameter $n$ and when $n\to \infty$ will the family of curves converge to a single curve? If so how to find it?
On
If $x>1$, $x^n$ will go to infinity for large $n$, so $1$ can be neglected. Therefore $f(x)=x$ for $x>1$. If $x\lt 1$, $x^n$ goes to zero, so it can be neglected. $f(x)=1$ for $x\lt 1$ The first search result in google is a graph of a function. Search for "plot (1-x^5)^0.2"
You want to find the value of the limit for each (fixed) $x$. For example, $f(0) = 1$. I don't want to give away the question entirely, but I suggest you use the 'binomial' expansion for general powers (see here, for example). Since you want a large-$n$ limit, you don't need to be really precise but can collect terms into $\mathcal{O}(1/n)$, or such. Note that you have to be careful to apply the $(1+z)^r$ expansion for $z < 1$. For $x > 1$, and hence $x^n > 1$, you should think about how to overcome this.
If you need a further hint, feel free to ask! :)