Equivalence of different definitions of continuity Let $f:\mathbb{R}\rightarrow\mathbb{R}$ be a real function. $f$ is continuous at point $c$ iff
$$(\forall\epsilon>0)(\exists\delta>0)(\forall x)(|x-c|<\delta\Rightarrow|f(x)-f(c)|<\epsilon)$$
Continuity is also defined at a point $c$ if $\lim\limits_{x\rightarrow c}f(x)=f(c)$ i.e.
$$(\forall\epsilon>0)(\exists\delta>0)(\forall x)(0<|x-c|<\delta\Rightarrow|f(x)-f(c)|<\epsilon)$$
So this is a trivial question but how do I show these definitions are equivalent? They are almost the same except that in the limit definition we have $0<|x-c|<\delta$ So how can I get rid of it in this case. As I said it is trivial and one can see it but can't see how to formally show it, so any help please.

Take the first definition as given. Then clearly the second condition follows, since the range of $x$ values being considered is a subset of the range provided by the first definition.
If we take the second definition as given, we have to show that the weaker condition $$0<|x-c|<\delta\implies |f(x)-f(c)|<\epsilon$$ implies the stronger one $$|x-c|<\delta\implies |f(x)-f(c)|<\epsilon$$ So we need only verify that for $x=c$, we still have $|f(x)-f(c)|<\epsilon$. But this is easy, because when $x=c$, $|f(x)-f(c)|=|f(c)-f(c)|=0<\epsilon$ regardless of the value of $\epsilon>0$.
