Proving continuity of a function composition on $\mathbb{R}^m \to \mathbb{R}^n \to \mathbb{R}^k$

Question

I wanted to proof that if $f\colon \mathbb{R}^m \to \mathbb{R}^n$ is continuous at $a\in \mathbb{R}^m$ and if $g\colon \mathbb{R}^n \to \mathbb{R}^k$ is continuous at $g(f(a))$, then $g\circ f$ is continuous at $a$.

I know that continuity of $f$ at $a$ implies $\forall \epsilon_1$ $\exists \delta_1$ such that if $\vert\vert x-a\vert\vert<\delta_1 \implies \vert\vert f(x)-f(a)\vert\vert<\epsilon_1.$ And also the continuity of $g$ at $f(a)$ implies $\forall \epsilon_2$ $\exists \delta_2$ such that if $\vert\vert y-f(a)\vert\vert<\delta_2 \implies \vert\vert g(y)-g(f(a))\vert\vert<\epsilon_2$.

Now, how do I properly combine this to prove that the composition is continuos at $a$, which implies proving $\forall \epsilon$ $\exists \delta$ such that if $\vert\vert x-a\vert\vert<\delta \implies \vert\vert (g\circ f)(x)-(g\circ f)(a)\vert\vert<\epsilon.$

Answer 1

For any $\varepsilon > 0$ we find $\delta' > 0$ such that $$\|f(x) - f(a)\| < \delta' \quad \rightarrow \quad \|(g \circ f)(x) - (g \circ f)(a)\| < \varepsilon$$ Furthermore, for this particular $\delta'$ we find $\delta > 0$ such that $$\|x - a\| < \delta \quad \rightarrow \quad \|f(x) - f(a)\| < \delta'$$ In conclusion, for $\varepsilon > 0$ we find $\delta > 0$ such that $$\|x - a\| < \delta \quad \rightarrow \quad \|f(x) - f(a)\| <\delta'\quad \rightarrow \quad \|(g \circ f)(x) - (g \circ a)(a)\| < \varepsilon$$