# Composition of multiple variable limits

I need help to prove the following theorem: Let $f,g$ be functions, $a \in D(f \circ g)$ a limit point. If $\lim_{x \to a} f(x)=b$ and $\lim_{y \to b} g(y)=c$, then $\lim_{x \to a} g(f(x)) = c$, if $g$ is continuous at b. Thanks in advance.

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There are a few things that need clarifying here: $a$ is a limit point of what? Does $D$ mean the domain? Does $\lim_{x \to a} f(x)$ really make sense (i.e. is $a$ also in the domain of $f$?) – Anthony Carapetis Sep 7 '13 at 4:24

Since $\lim_{x\to a}f(x)=b$ then for every $\delta>0$ there exists $\delta_{1}$ such that if $0<|x-a|<\delta_{1}$ then $|f(x)-b|<\delta$. Since $g$ is continuous at $b$ then for every $\epsilon>0$ there is $\delta>0$ such that if $|y-b|<\delta$ then $|g(y)-c|<\epsilon$. Choosing $0<|x-a|<\delta_{1}$ then $|f(x)-b|<\delta$. For all $x$ such that $0<|x-a|<\delta_{1}$ we have either $f(x)=b$ or $f(x)\neq b$ but $|f(x)-b|<\delta$. Since $g$ is continuous at $b$ then either case results in $|g(f(x))-c|<\epsilon$ since $|f(x)-b|<\delta$. $\epsilon$ was arbitrary so $\lim_{x\to a}g(f(x))=c$