Answer
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Work Step by Step
We say that $\lim_{x\to0}g(x)=k$ implies the limit of $g(x)$ as $x$ approaches $0$ is the number $k$, if, for every number $\epsilon\gt0$, there exists a corresponding number $\delta\gt0$ such that for all $x$, $0\lt |x| \lt \delta$, we get $|g(x)-k|\lt\epsilon$.
