Answer
We prove that $f\left( x \right) = \sin \left( {\dfrac{x}{2}} \right) + \cos \left( {\dfrac{x}{3}} \right)$ is represented by its Maclaurin series for all $x$.
Work Step by Step
From Table 2 we have
$\sin x = \mathop \sum \limits_{n = 0}^\infty \dfrac{{{{\left( { - 1} \right)}^n}{x^{2n + 1}}}}{{\left( {2n + 1} \right)!}} = x - \dfrac{{{x^3}}}{{3!}} + \dfrac{{{x^5}}}{{5!}} - \dfrac{{{x^7}}}{{7!}} + \cdot\cdot\cdot$, ${\ \ \ }$ converges for all $x$
$\cos x = \mathop \sum \limits_{n = 0}^\infty \dfrac{{{{\left( { - 1} \right)}^n}{x^{2n}}}}{{\left( {2n} \right)!}} = 1 - \dfrac{{{x^2}}}{{2!}} + \dfrac{{{x^4}}}{{4!}} - \dfrac{{{x^6}}}{{6!}} + \cdot\cdot\cdot$, ${\ \ \ }$ converges for all $x$
Therefore,
$\sin \left( {\dfrac{x}{2}} \right) = \mathop \sum \limits_{n = 0}^\infty \dfrac{{{{\left( { - 1} \right)}^n}{{\left( {\dfrac{x}{2}} \right)}^{2n + 1}}}}{{\left( {2n + 1} \right)!}} = \mathop \sum \limits_{n = 0}^\infty \dfrac{{{g^{\left( n \right)}}\left( 0 \right)}}{{\left( {2n + 1} \right)!}}{x^{2n + 1}}$, ${\ \ \ }$ where ${g^{\left( n \right)}}\left( 0 \right) = \dfrac{{{{\left( { - 1} \right)}^n}}}{{{2^{2n + 1}}}}$
$\cos \left( {\dfrac{x}{3}} \right) = \mathop \sum \limits_{n = 0}^\infty \dfrac{{{{\left( { - 1} \right)}^n}{{\left( {\dfrac{x}{3}} \right)}^{2n}}}}{{\left( {2n} \right)!}} = \mathop \sum \limits_{n = 0}^\infty \dfrac{{{h^{\left( n \right)}}\left( 0 \right)}}{{\left( {2n} \right)!}}{x^{2n}}$, ${\ \ \ }$ where ${h^{\left( n \right)}}\left( 0 \right) = \dfrac{{{{\left( { - 1} \right)}^n}}}{{{3^{2n}}}}$
So,
$f\left( x \right) = \sin \left( {\dfrac{x}{2}} \right) + \cos \left( {\dfrac{x}{3}} \right) = \mathop \sum \limits_{n = 0}^\infty \dfrac{{{g^{\left( n \right)}}\left( 0 \right)}}{{\left( {2n + 1} \right)!}}{x^{2n + 1}} + \mathop \sum \limits_{n = 0}^\infty \dfrac{{{h^{\left( n \right)}}\left( 0 \right)}}{{\left( {2n} \right)!}}{x^{2n}}$
Since $|{g^{\left( n \right)}}\left( 0 \right) = \dfrac{{{{\left( { - 1} \right)}^n}}}{{{2^{2n + 1}}}}| \le 1$ and $|{h^{\left( n \right)}}\left( 0 \right) = \dfrac{{{{\left( { - 1} \right)}^n}}}{{{3^{2n}}}}| \le 1$, we apply Theorem 2 by choosing $K=1$ such that
$|{g^{\left( n \right)}}\left( 0 \right) = \dfrac{{{{\left( { - 1} \right)}^n}}}{{{2^{2n + 1}}}}| \le K$ ${\ \ \ }$ and ${\ \ \ }$ $|{h^{\left( n \right)}}\left( 0 \right) = \dfrac{{{{\left( { - 1} \right)}^n}}}{{{3^{2n}}}}| \le K$,
for any value of $x$.
Therefore, By Theorem 2: $\sin \left( {\dfrac{x}{2}} \right)$ is represented by the Maclaurin series $\mathop \sum \limits_{n = 0}^\infty \dfrac{{{g^{\left( n \right)}}\left( 0 \right)}}{{\left( {2n + 1} \right)!}}{x^{2n + 1}}$, and $\cos \left( {\dfrac{x}{3}} \right)$ is represented by the Maclaurin series $\mathop \sum \limits_{n = 0}^\infty \dfrac{{{h^{\left( n \right)}}\left( 0 \right)}}{{\left( {2n} \right)!}}{x^{2n}}$.
Hence, $f\left( x \right) = \sin \left( {\dfrac{x}{2}} \right) + \cos \left( {\dfrac{x}{3}} \right)$ is represented by its Maclaurin series for all $x$.
