Give the \(\varepsilon\), \(\delta\) definition for continuity of a function \(f: \mathbb{R} \to \mathbb{R}\) at a point \(x_0 \in \mathbb{R}\).
Assume that \(f: \mathbb{R} \to \mathbb{R}\) is continuous and that \(\{a_n\}\) is a sequence of real numbers with \(\lim_{n \to \infty} a_n = L\). Prove that \(\lim_{n \to \infty} f(a_n) = f(L)\).
<1>2. Suppose \(a_n\to L\) and \(f\) is continuous at \(L\). Let \(\varepsilon>0\).
<1>3. There exists \(\delta>0\) such that \[ |x-L|<\delta\implies |f(x)-f(L)|<\varepsilon. \]
<1>4. There exists \(N\in\mathbb N\) such that \[ n\ge N\implies |a_n-L|<\delta. \]
<1>5. Hence, for every \(n\ge N\), \[ |f(a_n)-f(L)|<\varepsilon. \]
<1>6. Therefore \(f(a_n)\to f(L)\).
Solution
1. The function \(f\) is continuous at \(x_0\) if for every \(\varepsilon>0\) there exists \(\delta>0\) such that\[ |x-x_0|<\delta\implies |f(x)-f(x_0)|<\varepsilon. \]
This is continuity of \(f\) at \(L\) from 1.
This is the definition of \(a_n\to L\).
Apply 3 to \(x=a_n\), using 4.
The estimate in 5 is the definition of convergence of the sequence \(f(a_n)\) to \(f(L)\).