Answer
$$\lim_{x\to0}g(x)=k$$
According to the definition, this means for every arbitrary number $\epsilon\gt0$, there exists a corresponding number $\delta\gt0$ such that for all $x$,
$$0\lt|x-0|\lt\delta\Rightarrow|g(x)-k|\lt\epsilon$$
Work Step by Step
$$\lim_{x\to0}g(x)=k$$
According to the definition, this means for every arbitrary number $\epsilon\gt0$, there exists a corresponding number $\delta\gt0$ such that for all $x$,
$$0\lt|x-0|\lt\delta\Rightarrow|g(x)-k|\lt\epsilon$$
This, in other words, means that as we wish to get $\epsilon$ smaller and smaller (the value of $g(x)$ fluctuates less and less around $k$), we can always find a range around the point $x=0$ (which is $(-\delta,\delta)$) so that our goal can be achieved.
