Answer
There is nothing which can be said about the existence of $\lim_{x\to0}f(x)$.
Work Step by Step
$f(x)$ is defined for all $x$ in $[-1,1]$. Unfortunately, this is not enough to conclude anything about the existence of $\lim_{x\to0}f(x)$.
I will give examples of two cases where $\lim_{x\to0}f(x)$ exists and $\lim_{x\to0}f(x)$ does not exist. The graphs are shown below.
1) $f(x)=x$
- $f(x)$ is obviously defined for all $x$ in $[-1,1]$.
- To the left of $x=0$, $f(x)\to0$ as $x\to0$. To the right of $x=0$, $f(x)\to0$ as $x\to0$.
- Therefore, $\lim_{x\to0}f(x)$ does exist and equals $0$.
2) $f(x)=x+1$ for $x\lt0$ and $f(x)=x+2$ for $x\ge0$
- $f(x)$ is obviously defined for all $x$ in $[-1,1]$.
- To the left of $x=0$, $f(x)\to1$ as $x\to0$. To the right of $x=0$, $f(x)\to2$ as $x\to0$.
- This means as $x\to0$, there is not a single value approached by $f(x)$; its value jumps at $x=0$. So $\lim_{x\to0}f(x)$ does not exist in this case.