Show that the Taylor series for \(f(x)=e^x\), as given in Key Idea 32, is equal to \(f(x)\) by applying Theorem 77; that is show \(\lim\limits_{n\to\infty} R_n(x)=0\).
Per theorem 76, \(|R_n(x)| \le \frac{max|f^{n+1}(z)|}{(n+1)!} |x^{n+1}|\).
Derivative of \(e^x\) is \(e^x\), so \(|R_n(x)| \le \frac{e^z}{(n+1)!} |x^{n+1}|\).
For any \(x\), \(\lim\limits_{n\to\infty} \frac{e^z x^{n+1}}{(n+1)!} = 0\). That means that \(\lim\limits_{n\to\infty} R_n(x) = 0\).
Per theorem 77, \(f(x) = \sum\limits_{n=0}^{\infty} \frac{f^{(n)}(c)}{n!} (x-c)^n\).
Setting \(c=0\), \(f(x) = \sum\limits_{n=0}^{\infty} \frac{e^0}{n!} (x-0)^n = \sum\limits_{n=0}^{\infty} \frac{x^n}{n!} = e^x\), per Key Idea 32.