Suppose \(L\) is a real number and \(f, g: \mathbb{R} \to \mathbb{R}\). Prove that if \(\lim_{x \to 0} g(x) = L\) and \(f\) is continuous at \(L\), then \(\lim_{x \to 0} f(g(x))\) also exists.
<1>2. Let \(\varepsilon>0\). Since \(f\) is continuous at \(L\), there exists \(\eta>0\) such that \[ |y-L|<\eta\implies |f(y)-f(L)|<\varepsilon. \]
<1>3. Since \(g(x)\to L\) as \(x\to0\), there exists \(\delta>0\) such that \[ 0<|x|<\delta\implies |g(x)-L|<\eta. \]
<1>4. Therefore, whenever \(0<|x|<\delta\), \[ |f(g(x))-f(L)|<\varepsilon. \]
<1>5. Hence \(\lim_{x\to0}f(g(x))\) exists and equals \(f(L)\).
Solution
1. We claim that\[ \lim_{x\to0}f(g(x))=f(L). \]
By 3, the number \(g(x)\) satisfies \(|g(x)-L|<\eta\); applying 2 with \(y=g(x)\) yields the displayed inequality.
This is exactly the \(\varepsilon\)-\(\delta\) definition of the limit, using 4.