Numerical Value for $\lim \limits_{n \to \infty}\frac{x^n}{1+x^n}$ 
Let $$f (x) := \lim \limits_{n \to \infty}\frac{x^n}{1+x^n}$$
Determine the numerical value of $f(x)$ for all real numbers $x \ne -1$. For what values of $x$ is $f$ continuous?

I honestly do not know how to find the numerical value.  I don't even know what this means as the teachers notes do not seem to cover this topic directly.
For the second part the answer seems too obvious, so I feel like my thinking is off.
$f$ is continuous for all values of $x$ except for $x=-1$ and where $n$ is not odd.
Any assistance on this is greatly appreciated.
 A: Hint:
$$\frac{x^n}{1+x^n}=1-\frac1{1+x^n}$$
A: when $x > 1 $divide the numerator and denominator by $x^n$. As n tends to infinity 
$1/x^n$ tends to zero so the limit is 1.
when $0<x<1$ the numerator tends to zero and the denominator tend to 1. So the limit is 0.
A: For $|x|>1$, the limit is $1$ (rewrite the undeterminate as $\frac1{1+x^{-n}}$) and for $|x|<1$, it is $0$.
For $x=1$, $\frac12$ and for $x=-1$, undefined as the denominator can cancel out.
So the limit is piecewise constant, hence continous except at $x=1$.
A: $$\frac{x^n}{1+x^n}=\frac{\frac{x^n}{x^n}}{\frac{1}{x^n}+\frac{x^n}{x^n}}=\frac{1}{1+\frac{1}{x^n}}$$
A: In general you have for $f(x)=\frac{x^n+P(x)}{x^n+Q(x)}$ that $\lim \limits_{n \to \infty}f(x)=1$ if $P$ and $Q$ are polynomials both of degree less than $n$ (divide numerator and denominator by $x^n$ to see this).
With your function $f(x)=\frac{x^n}{x^n+1}$ you have no discontinuity when $n$ is even (because the denominator never is zero)  and just one discontinuity at $x=-1$ when $n$ is odd (because the denominator becomes zero).
