For a function \(f: \mathbb{R} \to \mathbb{R}\), give the definition of continuity of \(f\) at a point \(a\). For functions \(f\) and \(g\) from \(\mathbb{R}\) to \(\mathbb{R}\), prove that if \(f\) is continuous at \(a\) and \(g\) is continuous at \(b = f(a)\), then \(g \circ f\) is continuous at \(a\). [The composition \(g \circ f\) is defined by \((g \circ f)(x) = g(f(x))\) for \(x \in \mathbb{R}\).]
Solution
A function \(f:\mathbb R\to\mathbb R\) is continuous at \(a\) if for every \(\varepsilon>0\) there exists \(\delta>0\) such that \[ |x-a|<\delta\quad\Longrightarrow\quad |f(x)-f(a)|<\varepsilon. \]
Assume \(f\) is continuous at \(a\) and \(g\) is continuous at \(b=f(a)\). Let \(\varepsilon>0\). By continuity of \(g\) at \(b\), choose \(\eta>0\) such that \[ |y-b|<\eta\quad\Longrightarrow\quad |g(y)-g(b)|<\varepsilon. \] By continuity of \(f\) at \(a\), choose \(\delta>0\) such that \[ |x-a|<\delta\quad\Longrightarrow\quad |f(x)-f(a)|<\eta. \] Hence, if \(|x-a|<\delta\), then \[ |g(f(x))-g(f(a))|<\varepsilon. \] Therefore \(g\circ f\) is continuous at \(a\).