Suppose \(f, g: \mathbb{R} \to \mathbb{R}\) and \(a \in \mathbb{R}\).
State the \(\varepsilon\)-\(\delta\) definition of what it means for \(f\) to be continuous at \(a\).
Prove that if \(f\) and \(g\) are continuous at \(3\), then their sum \(f + g\) is also continuous at \(3\).
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 \(f\) and \(g\) are continuous at \(3\). Let \(\varepsilon>0\). Choose \(\delta_f>0\) so that \[ |x-3|<\delta_f\Longrightarrow |f(x)-f(3)|<\varepsilon/2, \] and choose \(\delta_g>0\) so that \[ |x-3|<\delta_g\Longrightarrow |g(x)-g(3)|<\varepsilon/2. \] Set \(\delta=\min\{\delta_f,\delta_g\}\). Then if \(|x-3|<\delta\), \[ |(f+g)(x)-(f+g)(3)| \le |f(x)-f(3)|+|g(x)-g(3)| <\varepsilon. \] Hence \(f+g\) is continuous at \(3\).