EXAMPLE 2 Prove that 9ex is equal to the sum of its Maclaurin series. SOLUTION If f(x) = 9ex, then f (n + 1)(x) = for all n. If d is any positive number and |x| ≤ d, then |f (n + 1)(x)| = ≤ 9ed. So Taylor's Inequality, with a = 0 and M = 9ed, says that |Rn(x)| ≤ (n + 1)! |x|n + 1 for |x| ≤ d. Notice that the same constant M = 9ed works for every value of n. But, from this equation, we have lim n → [infinity] 9ed (n + 1)! |x|n + 1 = 9ed lim n → [infinity] |x|n + 1 (n + 1)! = . It follows from the Squeeze Theorem that lim n → [infinity] |Rn(x)| = 0 and therefore lim n → [infinity] Rn(x) = for all values of x. By this theorem, 9ex is equal to the sum of its Maclaurin series, that is, 9ex = [infinity] 9xn n! n = 0 for all x.

If [tex]f(x) = 9e^x[/tex], then [tex]f ^{(n + 1)(x)} =9e^x[/tex] for all n. If d is any positive number and |x| ≤ d, then [tex]|f^{(n + 1)(x)}| = 9e^x\leq 9e^d.[/tex]

So Taylor's Inequality, with a = 0 and M = [tex]9e^d[/tex], says that [tex]|R_n(x)| \leq \dfrac{9e^d}{(n+1)!} |x|^{n + 1} \:for\: |x| \leq d.[/tex]

Notice that the same constant [tex]M = 9e^d[/tex] works for every value of n.

But, since [tex]lim_{n\to\infty}\dfrac{x^n}{n!} =0 $ for every real number x$[/tex],

We have [tex]lim_{n\to\infty} \dfrac{9e^d}{(n+1)!} |x|^{n + 1} =9e^d lim_{n\to\infty} \dfrac{|x|^{n + 1}}{(n+1)!} =0[/tex]

It follows from the Squeeze Theorem that [tex]lim_{n\to\infty} |R_n(x)|=0[/tex] and therefore [tex]lim_{n\to\infty} R_n(x)=0[/tex] for all values of x.

[tex]THEOREM\\If f(x)=T_n(x)+R_n(x), $where $T_n $is the nth degree Taylor Polynomial of f at a and $ lim_{n\to\infty} R_n(x)=0 \: for \: |x-a|<R, $then f is equal to the sum of its Taylor series on $ |x-a|<R[/tex]

By this theorem above, [tex]9e^x[/tex] is equal to the sum of its Maclaurin series, that is,

[tex]9e^x=\sum_{n=0}^{\infty}\frac{9x^n}{n!}[/tex] for all x.