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 \(\lim\limits_{n\to\infty} R_n(x)=0\).
\(|R_n(x)| \le \frac{max|f^{n+1}(z)|}{(n+1)!} |x^{n+1}|\).
\(|R_n(x)| \le \frac{e^z}{(n+1)!} |x^{n+1}|\)
For any \(x\), \(\lim\limits_{n\to\infty} \frac{e^z x^{n+1}}{(n+1)!} = 0\).
\(\lim\limits_{n\to\infty} R_n(x) = 0\).
\(f(x) = \sum\limits_{n=0}^{\infty} \frac{f^{(n)}(c)}{n!} (x-c)^n\).
If \(c=0\)
\(f(x) = \sum\limits_{n=0}^{\infty} \frac{e^0}{n!} (x-0)^n = \sum\limits_{n=0}^{\infty} \frac{x^n}{n!} = e^x\)